





















































































N 


I 




i 


ENCYCLOPEDIA 

= if =~ 

CARPENTRY 
and BUILDING 

A Complete Manual of Carpentry and Joinery, 
Covering Every Known Detail of the Trade. 
Designed to Afford Practical Help in the 
Every-Day Problems of the Car¬ 
penter, the Builder and 
the Architect 


Arranged anji Edited by 

FREiyhY HODGSON 

Architect 

/ - 

ILLUSTRATED WITH OVER 
3000 ENGRAVINGS 


Complete in Four Volumes 


CHICAGO 

NATIONAL INSTITUTE OF PRACTICAL MECHANICS 

PUBLISHERS 

















Copyright 1916 and 1914 

BY 

NATIONAL INSTITUTE OF PRACTICAL MECHANICS 

< 



FEB 21 1916 

©CU42''!M8 


'Hv-t 


t 


MODERN CARPENTRY 

PART I 

CARPENTER’S GEOMETRY 
CHAPTER I 

THE CIRCLE 

While it is not absolutely necessary that, to become 
a good mechanic, a man must need be a good scholar 
or be well advanced in mathematics or geometry, yet, 
if a man be proficient in these sciences they will be a 
great help to him in aiding him to accomplish his work 
with greater speed and more exactness than if he did 
not know anything about them. This, I think, all will 
admit. It may be added, however, that a man, the 
moment he begins active operations in any of the con¬ 
structional trades, commences, without knowing it, to 
learn the sciene^ of geometry in its rudimentary 
stages. He wishes to square over a board and employs 
a steel or other square for this purpose, and, when he 
scratches or pencils a line across the board, using the 
edge or the tongue of the square as a guide, while the 
edge of the blade is against the edge of the board or 
parallel with it, he thus solves his first geometrical 
problem, that is, he makes a right angle with the edge 
of the board. This is one step forward in the path of 
geometrical science. 

He desires to describe a circle, say of eight inches 
diameter. He knows instinctively that if he opens his 

9 


10 


MODERN CARPENTRY 


compasses until the points of the legs are four inches 
apart,—or making the radius four inches—he can, by 
keeping one point fixed, called a “center,” describe a 
circle with the other leg, the diameter of which will 
be eight inches. By this process he has solved a 
second geometrical problem, or at least he has solved 
it so far that it suits his present purposes. These 
examples, of course, do not convey to the operator the 
more subtle qualities of the right angle or the circle, 
yet they serve, in a practical manner, as assistants in 
every-day work. 

When a man becomes a good workman, it goes with¬ 
out saying that he has also become possessor of a fair 
amount of practical geometrical knowledge, though he 
may not be aware of the fact. 

The workman who can construct a roof, hipped, 
gabled, or otherwise, cutting all his material on the 
ground, has attained an advanced practical knowledge 
of geometry, though he may never have heard of 
Euclid or opened a book relating to the science. 
Some of the best workmen I have met were men who 
knew nothing of geometry as taught in the books, yet 
it was no trouble for them to lay out a circular or 
elliptical stairway, or construct a rail over them, a 
feat that requires a knowledge of geometry of a high 
order to properly accomplish. 

These few introductory remarks are made with the 
hope that the reader of this little volume will not be 
disheartened at the threshold of his trade, because of 
his lack of knowledge in any branch thereof. To 
become a good carpenter or a good joiner, a young 
man must begin at the bottom, and first learn his 
A, B, C’s, and the difficulties that beset him will disap¬ 
pear one after another as his lessons are learned. It 


4 


CARPENTER’S GEOMETRY 


11 


must always be borne in mind, however, that the young 
fellow who enters a shop, fully equipped with a knowl¬ 
edge of general mathematics and geometry, is in a 
much better position to solve the work problems that 
crop up daily, than the one who starts work without 
such equipment. If, however, the latter fellow be a 
boy possessed of courage and perseverance, there is no 



reason why he should not “catch up”—even over¬ 
take—the boy with the initial advantages, for what is 
then learned will be more apt to be better understood, 
and more readily applied to the requirements of his 
work. To assist him in “catching up” with his more 
favored shopmate, I propose to submit for his benefit 
a brief description and explanation of what may be 
termed “Carpenter’s Geometry/’ which will be quite 






12 


MODERN CARPENTRY 


sufficient if he learn it well, to enable him to execute 
any work that he may be called upon to perform; and 
I will do so as clearly and plainly as possible, and in 
as few words as the instructions can be framed so as to 
make them intelligible to the student. 

The circle shown in Fig. i is drawn from the center 
2, as shown, and may be said to be a plain figure 
within a continual curved line, every part of the line 
being equally distant from the center 2. It is the 
simplest of all figures to draw. The line AB, which 
cuts the circumference, is called the diameter, and the 
line DE is denominated a chord, and the area en¬ 
closed within the curved line and the chord is termed 
a segment. The radius of a circle is a line drawn from 
the center 2 to the circumference C, and is always one- 
half the length of the diameter, no matter what that 
diameter may be. A tangent is a line which touches 
the circumference at some point and is at right angles 
with a radial line drawn to that point as shown at C. 

The reader should remember these definitions as 
they will be frequently used when explanations of 
other figures are made; and it is essential that the 
learner should memorize both the terms and their sig¬ 
nifications in order that he may the more readily 
understand the problems submitted for solution. 

It frequently happens that the center of a circle is 
not visible but must be found in order to complete 
the circle or form some part of the circumference. 
The center of any circle may be found as follows: let 
BHC, Fig. 2 , be a chord of the segment H; and 
BJA a chord enclosing the segment. Bisect or 
divide in equal parts, the chord BC, at H, and square 
down from this point to D. Do the same with the 
chord AJB, squaring over from J to D, then the 


CARPENTER’S GEOMETRY 


i3 


point where JD and HD intersect, will be the center 
of the circle. 

This is one of the most important problems for the 
carpenter in the whole range of geometry as it enables 
the workman to locate any center, and to draw curves 
he could not otherwise describe without this or other 
similar methods. It is by aid of this problem that 
through any three points not in a straight line, a 



circle can be drawn that will pass through each of the 
three points. Its usefulness will be shown further on 
as applied to laying out segmental or curved top 
window, door and other frames and sashes, and the 
learner should thoroughly master this problem before 
stepping further, as a full knowledge of it will assist 
him very materially in understanding other problems. 

The circumference of every circle is measured by 
being supposed to be divided into 360 equal parts, 
called' degrees', each degree containing 60 minutes , a 



14 


MODERN CARPENTRY 


smaller division, and each minute into 60 seconds , a 
still smaller division. Degrees, minutes, and seconds 
are written thus: 45 0 15' 30", which is read, forty-five 
degrees, fifteen minutes, and thirty seconds. This, I 
think, will be quite clear to the reader. Arcs are meas¬ 
ured by the number of degrees which they contain: thus, 
in Fig. 3, the arc AE, which contains 90°, is called a 
quadrant, or the quarter of a circumference, because 



90° is one quarter of 360°, and the arc ABC whicV* eon- 
tains 180 0 , is a semi-circumference. Every angle is also 
measured by degrees, the degrees being reckoned on an 
arc included between its sides; described from the ver¬ 
tex of the angle as a center, as the point O, Fig. 3; 
thus, AOE contains 90°; and the angle BOD, which is 
half a right angle, is called an angle of 45°, which is 







CARPENTER’S GEOMETRY 




the number it contains, as will be seen by counting 
off the spaces as shown by the divisions on the curved 
line BD. These rules hold good, no matter what may 
be the diameter of the circle. If large, the divisions 
are large; if small, the divisions are small, but the 
manner of reckoning is always the same. 

One of the qualities of the circle is, that when 
divided in two by a diameter, making two semicircles, 
any chord starting at the extremity of such a diameter, 
as at A or B, Fig. 4, and cutting the circumference at 
any point, as at C, D or E, a line drawn from this 



point to the other extremity of the diameter, will 
form a right angle—or be square with the first chord, 
as is shown by the dotted lines BCA, BDA, and BEA. 
This is something to be remembered, as the problem 
will be found useful on many occasions. 

The diagram shown at Fig. 5 represents a hexagon 
within a circle. This is obtained by stepping around 
the circumference, with the radius of the circle on the 
compasses, six times, which divides the circumference 
into six equal parts; then draw lines to each point, 
which, when completed, will form a hexagon, a six- 
sided figure. By drawing lines from the points 
obtained in the circumference to the center, we get a 




MODERN CARPENTRY 


16 

three-sided figure, which is called an equilateral tri¬ 
angle, that is, a triangle having all its sides equal in 



length; as AB, AC and BC. The dotted lines show how 
an equilateral triangle may be produced on a straight 
line if necessary. 

The diagram shown 
at Fig. 6 illustrates the 
method of trisecting a 
right angle or quadrant 
into three equal parts. 

Let A be a center, and 
with the same radius 
intersect at E, thus the 
quadrant or right angle 
is divided into three 
equal parts. 












CARPENTER’S GEOMETRY 


*7 


If we wish to get the length of a straight line that 
shall equal the circumference of a circle or part of 
circle or quadrant, we can do so by proceeding as fol¬ 
lows: Suppose Fig. 7 to represent half of the circle, 
as at ABC; then draw the chord BC, divide it at P, 
join it at A; then four times PA is equal to the cir¬ 
cumference of a circle whose diameter is AC, or equal 
to the curve CB. 

To divide the quadrant AB into any number of 
equal parts, say thirteen, we simply lay on a rule and 
make the distance from A to R measure three and one- 



fourth inches, which are thirteen quarters or parts on 
the rule; make R2 equal one-fourth of an inch; join 
RP; draw from 2 parallel with RP, cutting at V; now 
take PV in the dividers and set off from A on the circle 
thirteen parts, which end at B, each part being equal 
to PV, and the problem is solved. The “stretchout” 
or length of any curved line in the circle can then be 
obtained by breaking it into segments by chords, as 
shown at BN. 

I have shown in Fig. 5, how to construct an equi¬ 
lateral triangle by the use of the compasses. I give at 



i8 


MODERN CARPENTRY 


Fig 8 a practical example of how this figure, in con¬ 
nection with circles, may be employed in describing a 
figure known as the trefoil, a figure made much use of 
in the construction of church or other Gothic work and 
for windows and carvings on doors and panelings. 
Each corner of the triangle, as ABC, is a center from 
which are described the curves shown within the outer 
circles. The latter curves are struck from the center 



O, which is found by dividing the sides of the equi¬ 
lateral triangle and squaring down until the lines cross 
at O. The joint lines shown are the proper ones to 
be made use of by the carpenter when executing his 
work. The construction of this figure is quite simple 
and easy to understand, so that anyone knowing how to 
handle a rule and compass should be able to construct 
it after a few minutes’ thought. This figure is the key 
to most Gothic ornamentation, and is worth mastering. 








CARPENTER’S GEOMETRY 


19 


There is another method of finding the length or 
“stretchout” of the circumference of a circle, which I 
show herewith at Fig. 9. Draw the semicircle SZT, 
and parallel to the diameter ST draw the tangent UZV; 
upon S and T as centers, with ST as radius, mark the 
arcs TR and SR; from R, the intersection of the arcs, 
draw RS and continue to U; also draw RT, and con¬ 
tinue to V; then the line VU will nearly equal in 



length the circumference of the semicircle. The 
length of any portion of a circle may be found as fol¬ 
lows: Through X draw RW, then WU will be the 
“stretchout” or length of that portion of the circle 
marked SX. There are several other ways of deter¬ 
mining by lines a near approach to the length of the 
circumference or a portion thereof; but, theoretically, 
the exact “stretchout” of a circumference has not been 
found by any of the known methods, either arith- 




ao 


MODERN CARPENTRY 


metically or geometrically, though for all practical 
purposes the methods given are quite near enough. 
No method, however, that is given geometrically is so 
simple, so convenient and so accurate as the arith¬ 
metical one, which I give herewith. If we multiply 
the diameter of a circle by 3.1416, the product will 
give the 'length of the circumference, very nearly. 
These figures are based on the fact that a circle 
whose diameter is 1—say one yard, one foot, or one 
inch—will have a circumference of nearly 3.1416 times 
the diameter. 

With the exception 
of the formation of 
mouldings, and orna¬ 
mentation where the 
circle and its parts 
take a prominent 
part, I have sub¬ 
mitted nearly all con¬ 
cerning the figure, 
the everyday carpen¬ 
ter will be called 
upon to employ, and when I approach the chapter on 
Practical Carpentry later on, I will try and show how 
to use the knowledge now given. 

Before leaving the subject, however, it may be as 
well to show how a curve, having any reasonable 
radius, may be obtained—practically—if but three 
points in the circumference are available; as referred 
to in the explanation given of Fig. 5. Let us suppose 
there are three points given in the circumference of a 
circle, as ABC, Fig. 10, then the center of such circle 
can be found by connecting the points AB and BC 
by straight lines as shown, and by dividing these lines 



CARPENTER’S GEOMETRY 


91 


and squaring down as shown until the lines intersect at 
O as shown. This point O is the center of the circle. 

It frequently happens that it is not possible to find a 
place to locate a center, because of the diameter being 
so great, as in segmental windows and doors of large 
dimensions. To overcome this difficulty a method 



has been devised by which the curve may be correctly 
drawn by nailing three wooden strips together so as to 
form a triangle, as shown in Fig. II. Suppose NO to 
be the chord or width of frame, and QP the height of 
segment, measuring from the springing lines N and O; 
drive nails or pins at O and N, keep the triangle close 
against the nails, and place a pencil at P, then slide 
the triangle against the pins or nails while sliding, and 
the pencil will describe the necessary curve. The 
arms of the triangle should be several inches longer 
than the line NO, so that when the pencil P arrives at 
N or O, the arms will still rest against the pins, 





CHAPTER II 


POLYGONS 

A polygon is a figure that is bounded by any number 
of straight lines; three lines being the least that can 
be employed in surrounding any figure, as a triangle, 
Fig. i. 

A polygon having three sides is called a trigon; it is 
also called an equilateral triangle. A polygon of four 
sides is call a tetragon; it is also called a square and 

an equilateral rect¬ 
angle. A polygon 
of five sides is a 
pentagon. A poly¬ 
gon of six sides is a 
hexagon. A poly¬ 
gon of seven sides is 
called a heptagon. 
A polygon of eight 
sides is called an 
octagon. A polygon of nine sides is called a nonagon. 
A polygon of ten sides is called a decagon. A polygon 
of eleven sides is called an undecagon. And a poly¬ 
gon of twelve sides is called a dodecagon. 

There are regular and irregular polygons. Those 
having equal sides are regular; those having unequal 
sides are irregular. Polygons having more than twelve 
sides are known among carpenters by being denom¬ 
inated as a polygon having “so many sides/’ as a 
“polygon with fourteen sides,” and so on. 

22 




CARPENTER’S GEOMETRY 


23 


Polygons are often made use of in carpenter work, 
particularly in the formation of bay-windows, oriels, 
towers, spires, and similar work; particularly is this 
the case with the hexagon and the octagon; but the 
most used is the equilateral rectangle, or square; 
therefore it is essential that the carpenter should 
know considerable regarding these figures, both as to 
their qualities and their construction. 

The polygon having the least lines is the trigon, 
a three-sided figure. This is constructed as follows: 
Let CD, Fig. 1, be any given line, and the dis¬ 
tance CD the length of the side required. Then with 
one leg of the compass on D as a center, and the other 
on C, describe the arc 

shown at P. Then with JP _ \U^. 

C as a center, describe an¬ 
other arc at P, cutting the 
first arc. From this point 
of intersection draw the 
lines PD and PC, and the 
figure is complete. To 
get the miter joint of this 
figure, divide one side jp 
into two equal parts, and 

from the point obtained draw a line through opposite 
angle as shown by the dotted line, and this line will be 
the line of joint at C, or for any of the other angles. 

The square, or equilateral rectangle, Fig. 2, may be 
obtained by a number of methods, many of which will 
suggest themselves to the reader. I give one method 
that may prove suggestive. Suppose two sides of a 
square are given, LHN, the other sides are found by 
taking HL as radius, and with LN for centers make 
the intersection in P, draw LP and NP, which com- 







MODERN CARPENTRY 


pletes the figure. The miter for the joints of a figure 
of this kind is an angle of 45 0 , or the regular miter. 
The dotted line shows the line of “cut” or miter. 



To construct a pentagon we proceed as follows: 
Let AB, Fig. 3, be a given line and spaced off to the 
length of one side of the figure required; divide this 
line into two equal parts. From B square up a line; 

make BN equal to 
AB, strike an arc 
3N as shown by the 
dotted lines, with 2 
as a center and N 
as a radius, cutting 
the given line at 3. 
Take A3 for radius; 
from A and B as 
centers, make the 
intersection in D; 
from D, with 









CARPENTER’S GEOMETRY 


*5 


radius equal to AB, strike an arc; with the same radius 
and A and B as centers, intersect the arc in EC. By- 
joining these points the pentagon is formed. The cut, 
or angle of joints, is found by raising a line from 2 and 
cutting D, as shown by the dotted line. 

The hexagon, a six-sided figure shown at Fig. 4, is 
one of the simplest to construct. A quick method is 
described in Chapter I, when dealing with circles, but 
I show the method of construction in order to be cer¬ 
tain that the student may be the better equipped to 
deal with the figure. Take the length of one side of 
the figure on compasses; make this length the radius 
of a circle, thus describe a circle as shown. Start 
from any point, as at A, and step around the circum¬ 
ference of the circle with the radius of it, and the 
points from 
which to draw 
the sides are 
found, as the 
radius of any cir¬ 
cle will divide 
the circu in¬ 
ference of that 
circle into s i x 
equal parts. 

This figure may 
be drawn without 
first making a circle if necessary. Set off two equal 
parts, ABC, Fig. 5, making three centers; from each, 
with radius AC, make the intersection as shown, 
through which draw straight lines, and a hexagon is 
formed. The miter joint follows either of the straight 
lines passing through the center, the bevel indicating 
the proper angle. 






2 6 


MODERN CARPENTRY 


The construction of a heptagon or seven-sided 
figure may be accomplished as follows: Let AB, Fig. 
6, be a given line, and the distance AB the length of 
the side of the figure. Divide at K, square up from 
this point, then take AB for radius and B as a center; 
intersect the line from K at L; with same radius and 
A as center, draw the curve 2, 3; then take KL as 
radius, and from 2 as a center, intersect the circle at 3; 
draw from it to B, cutting at N, through which point 
draw from A; make AD equal B3; join A3 and BD; 
draw from 3 parallel with AD; draw from B through 
L, cutting at C; join it and A; draw from 3 parallel 

with AC; make 
3H equal AB, 
and CE equal 

ND; join ED; 

draw from H 
parallel with 3C, 
cutting at F, 
join this line 

and E, which 

completes the 
heptagon. It is 
not often this 
figure is used in 
carpentry, 
though I have sometimes employed it in constructing 
bay windows and dormers, using the four sides, 3H, 

HF, FE, and ED This makes a bold front, and 

serves well in a conservatory or other similar place. 

It is proper that the reader should know how to 
construct this figure, as it serves as an exercise, and 
illustrates a principle of drawing by parallels, a knowl¬ 
edge of which would be found invaluable to the ambi- 






CARPENTER’S GEOMETRY 


27 


tious young carpenter, who desires to become, not only 
a good workman, but a good draftsman as well. 

The octagon or eight-sided figure claims rank next 
to the square and circle, in point of usefulness to the 
general carpenter, owing partly to its symmetry of 
form, and its simplicity of construction. There are a 
great number of methods of constructing this figure, 
but I will give only a few of the simplest, and the ones 
most likely to be readily understood by the ordinary 
'workman. 

One of the simplest methods of forming an octagon 
is shown at Fig. 7, 
where the corners of 
the square are used as 
centers, and to the cen¬ 
ter A of the square for 
radius. Parts of a cir¬ 
cle are then drawn and 
continued until the 
boundary lines are cut. 

At the points found 
draw diagonal lines 
across the corner as 
shown, and the figure 
will be a complete octagon, having all its sides of 
equal length. 

The method of obtaining the joint cut or miter for 
an octagon is shown at Fig. 8, where the angle ABC, 
is divided into two equal angles by the following 
process: From B, with any radius, strike an arc, giving 
A and C as centers, from which, with any radius, make 
an intersection, as shown, and through it from B, draw 
a line, and the proper angle for the cut is obtained, the 
dotted line being the angle sought. By this method 




MODERN CARPENTRY 


28 

of bisecting an aqjSo 0 no matter how obtuse or acute it 
may be, the mitci? Joar£ or cut may be obtained. This 
x is a very useful prob- 

\ lem, as it is often 

called into requisi¬ 
tion for cutting 
mouldings in panels 
and other work, 
where the angles are 
not square, as in 
stair spandrils and 
raking wainscot. 

To construct an 
octagon when the length of one of its sides is given, 
as AB, Fig. 9, square up the two lines, AN, BF, then 

2T F 



take AB as radius with A and B as centers, and 
draw the arcs, cutting the two lines at C and J; 







CARPENTER’S GEOMETRY 


29 


draw from AB, through CJ, and again from A draw 
parallel with BJ; then draw from B parallel with 
AC; make BV and CF equal AB; join EV; make 
CF equal CA; square over FN; join FE; draw NP 
parallel with AC, then join PR, and the figure is 
complete. 



As the sides of all regular octagons are at an angle 
of 45 0 with each other, it follows that an octagon may 
be readily constructed by making use of a set square 
having its third side to correspond with an angle of 
45 0 , for by extending the line AB, and laying the set 
square on the line with one point at B, as shown in 
Fig. 10, the line BV, Fig. 9, can be drawn, and when 
made the same length as BV, the process can be 
repeated to VE; and so on until all the points have 
been connected. 

Suppose we have a square stick of timber 12 x 12 
inches, and any length, and we wish to make it an octa¬ 
gon; we will first be obliged to find the gauge points 
so as to mark the stick, to snap a chalk line on it so as 
to tell how much of the corners must be removed in 
order to give to the stick eight sides of equal width. 
We do this as follows: Make a drawing the size of a 










30 


MODERN CARPENTRY 


section ot the timber, that is, twelve inches square, 
then draw a line from corner to corner as AB, Fig. n, 
and make AC equal in length to AD, which is twelve 
inches; square over from C to K; set your gauge to 
BK, and run your lines to this gauge, and remove the 
corners off to lines, and the stick will then be an octa¬ 
gon having eight equal sides. 

There are a number of other methods of finding the 


B 



gauge points, some of which I may describe further on, 
but I think I have dwelt long enough on polygons to 
enable the reader to lay off all the examples given. 
The polygons not described are so seldom made use 
of in carpentry, that no authority that I am aware of 
describes them when writing for the practical work¬ 
man; though in nearly all works on theoretical geom- 





CARPENTER’S GEOMETRY 


3 i 


etry the figures are given with all their qualities. If 
the solution of any of the problems offered in this 
work requires a description and explanation of poly¬ 
gons with a greater number of sides than eight, such 
explanation will be given. 


CHAPTER III 


SOME STRAIGHT LINE SOLUTIONS 

The greatest number of difficult problems in carpen¬ 
try are susceptible of solution by the use of straight 
lines and a proper application of the steel square, and 
jp in this chapter I will 

endeavor to show the 
reader how some of 
the problems may be 
solved, though it is 
not intended to offer 
a treatise on the 
subject of the utility 
of the steel square, 
as that subject has 
been treated at 
length in other 
works, and another and exhaustive work is now in 
preparation; but it is thought no work on carpentry 
can be complete without, at least, showing some of the 
solutions that may be accomplished by the proper use 
of this wonderful instrument, and this will be done as 
we proceed. 

One of the most useful problems is one that enables 
us to make a perpendicular line on any given straight 
line without the aid of a square. This is obtained as 
follows: Let JK, Fig. i, be the given straight line, 
and make F any point in the square or perpendicular 
line required. From F with any radius, strike the arc 
32 







CARPENTER’S GEOMETRY 


33 


cutting in JK; with these points as centers, and any 
radius greater than half JK, make intersection as 
shown, and from this point draw a line to F, and this 
line is the perpendicular required. Foundations, and 
other works on a large scale are often “squared” or 
laid out by this method, or by another, which I will 
submit later. 

In a previous illustration I showed how to bisect an 
angle by using the compasses and straight lines, so as 
to obtain the proper joints or miters for the angles. At 
Fig. 2 ,1 show how this may be done by the aid of the steel 
square alone, 
as follows: The 
angle is ob¬ 
tuse, and may 
be that of an 
octagon or 
pentagon o r 
other polygon. 

Mark any two 
points on the 
angle, as DN, 
equally distant from the point of angle L; apply the 
steel square as shown, keeping the distance EN and 
ED the same, then a line running through the angle L 
and the point of the square E will be the line sought. 

To bisect an acute angle by the same method, pro¬ 
ceed as follows: Mark any two points AC, Fig. 3, 
equally distant from B; apply the steel square as 
shown, keeping its sides on AC; then the distance on 
each side of the square being equal from the corner 
gives it for a point, through which draw a line from B, 
and the angle is divided. Both angles shown are 
divided by the same method, making the intersection 


>L 

4 

/ 

/ 

/ 

/ 





34 


MODERN CARPENTRY 


in P the center of the triangle. The main thing to be 
considered in this solution is to have the distances A 

and C equal 
from the point 
B; also an 
equal distance 
from the point 
or toe of the 
square to the 
points of con¬ 
tact C and A 
on the boun¬ 
dary lines. 

A repetition 
of the same 
method of bisecting angles, under other conditions, is 
shown at Fig. 4. The process is just the same, and the 




reference letters are also the same, so any further 
explanation is unnecessary. 



CARPENTER’S GEOMETRY 


35 


To get a correct miter cut, or, in other words, an 
angle of 45°, on a board, make either of the points A 
or C, Fig. 5, the starting point for the miter, on the 
edge of the 
board, then ap¬ 
ply the square 
as shown, keep¬ 
ing the figure 
12" at A or C, 
as the case may 
be, with the fig¬ 
ure 12" on the 
other blade of the square on the edge of the board as 
shown; then the slopes on the edge of the square from 
A to B and C to B, will form angles of 45 0 with the 
base line AC. This problem is useful from many 
points of view, and will often suggest itself to the 
workman in his daily labor. 

To construct a figure showing on one side an angle 
of 30° and on the other an angle of 6o°, by the use of 


D 




the steel square, we go to work as follows: Mark on the 
edge of a board two equal spaces as AB, BC, Fig. 6, 
apply the square, keeping its blade on AC and making 






3* 


MODERN CARPENTRY 


AD equal AB; then the angles 30° and 60 0 are 
formed as shown. If we make a templet cut exactly 
as shown in Fig. 5, also a templet cut as shown in 
this last figure, and these templets are made of some 
hard wood, we get a pair of set squares for drawing 
purposes, by which a large number of geometrical 
problems and drawing kinks may be wrought out. 

The diameter of any circle within the range of the 
steel square may be determined by the instrument as 
follows: The corner of the square touching any part 
of the circumference A, Fig. 7, and the blade cutting 
in points C, B, gives the diameter of the circle as 



shown. Another application of this principle is, that 
the diameter of a circle being known, the square may 
be employed to describe the circumference. Suppose 
CB to be the known diameter; then put in two nails 
as shown, one at B and the other at C, apply the 
square, keeping its edges firmly against the nails, con¬ 
tinually sliding it around, then the point of the square 
A will describe half the circumference. Apply the 


CARPENTER’S GEOMETRY 


37 


square to the other side of the nails, and repeat the 
process, when the whole circle will be described. This 
problem may be applied to the solution of many others 
of a similar nature. 

At Fig. 8, I show how an equilateral triangle may 
be obtained by the use of a square. Draw the line 



DC; take 12 on the blade and 7 on the tongue; mark 
on the tongue for one side of the figure. Make the dis¬ 
tance from D to A equal to the desired length of one side 
of the figure. Reverse the square, placing it as shown by 
the dotted lines in the sketch, bringing 7 of the tongue 
against the point A. Scribe along the tongue, pro¬ 
ducing the line until it intersects the first line drawn 
in the point E, then AEB will be an equilateral tri¬ 
angle. A method of describing a hexagon by the 
square, is shown at Fig. 9, which is quite simple. 
Draw the line GH; lay off the required length of one 
side on this line, as DE. Place the square as before, 
with 12 of the blade and 7 of the tongue against the 
line GH; placing 7 of the tongue against the point D, 
scribe along the tongue for the side DC. Place the 
square as shown by the dotted lines; bringing 7 of the 
tongue against the point E, scribe the side EF. Con- 



3* 


MODERN CARPENTRY 


tinue in this way until the other half of the figure is 
drawn. All is shown by FABC. 

The manner of bisecting angles has been shown in 
Figs. 2, 3 and 4 of the present chapter, so that it is 
not necessary to repeat the process at this time. 

The method of describing an octagon by using 
the square, is shown at Fig. 10. Lay off a square 



section with any length of sides, as AB. Bisect this 
side and place the square as shown on the side 

AB, with the length 
bisected on the blade 
and tongue; then 
the tongue cuts the 
side at the point to 
gauge for the piece 
to be removed. To 
find the size of 
square required for 
an octagonal prism, 
when the side is 
given: Let CD equal 
the given side; place 
the square on the 














CARPENTER'S GEOMETRY 


39 


line of the side, with one-half of the side on the blade 
and tongue; then the tongue cuts the line at the point 
JB, which determines the size of the square, and the 
piece to be removed. 

A near approxima¬ 
tion to the length or 
stretch-out of a cir¬ 
cumference of a cir¬ 
cle may be obtained 
by the aid of the 
steel square and a 
straight line, as fol¬ 
lows: Take three 

diameters of the 

circle and measure up the side of the blade of the 
square, as shown at Fig. II, and fifteen-sixteenths of 
one diameter on the tongue. From these two points 









draw a diagonal, and the length of this diagonal will 
be the length or stretch-out of the circumference nearly. 

If it is desired to divide a board or other substance 
into any given number of equal parts, without going 
through the process of calculation, it may readily be 
done by the aid of the square or even a pocket rule. 
Let AC, BD, Fig. 12, be the width of the board or 

























40 


MODERN CARPENTRY 


other material, and this width is seven and one-quarter 
inches, and we wish to divide it into eight equal 
parts. Lay on the board diagonally, with furthermost 
point of the square fair with one edge, and the mark 
8 on the square on the other edge; then prick off the 
inches, I, 2 , 3, 4, 5, 6 and 7 as shown, and these points 
will be the gauge points from which to draw the 
parallel lines. These lines, of course, will be some¬ 
thing less than one inch apart. 

If the board should be more than eight inches wide, 
then a greater length of the square may be used, as 
for instance, if the board is ten inches wide, and we 
wish to divide it into eight equal parts, we simply 
make use of the figure 12 on the square instead of 8, 
and prick off the spaces every one and a half inches 
on the square. If the board is more than 12 inches 
wide, and we require the same number of divisions, we 
make use of figure 16 on the square, and prick off at 
every two inches. Any other divisions of the board 
may be obtained in a like manner, varying only the 
use of the figures on the square to get the number 
of divisions required. 

As a number of problems in connection with actual 
work, will be wrought out on similar lines to the fore¬ 
going, further on in this book, I will close this chapter 
in order to give as much space as possible in describ¬ 
ing the ellipse and the higher curves. 


CHAPTER IV 


ELLIPSES, SPIRALS, AND OTHER CURVES 

The ellipse, next to the circle, is the curve the car¬ 
penter will be confronted with more than any other, 
and while it is not intended to discuss all, or even a 
major part, of the properties and characteristics of 
this curve, I will endeavor to lay before the reader 
all in connection with it that he may be called upon 
to deal with. 

According to geometricians, an ellipse is a conic 
section formed by cutting a cone through the curved 
surface, neither parallel to the base nor making a 
subcontrary section, so that the ellipse like the circle 
is a curve that returns within itself, and completely 
encloses a space. One of the principal and useful 
properties of the ellipse is, that the rectangle under 
the two segments of a diameter is as the square of the 
ordinate. In the circle, the same ratio obtains, but 
the rectangle under the two segments of the diameter 
becomes equal to the square of the ordinate. 

It is not necessary that we enter into a learned 
description of the relations of the ellipse to the cone 
and the cylinder, as the ordinary carpenter may never 
have any practical use of such knowledge, though, if 
he have time and inclination, such knowledge would 
avail him much and tend to broaden his ideas. 
Suffice for us to show the various methods by which 
this curve may be obtained, and a few of its applica¬ 
tions to actual work. 

One of the simplest and most correct methods of 
describing an ellipse, is by the aid of two pins, a string 

41 


42 


MODERN CARPENTRY 


and a lead-pencil, as shown at Fig. I. Let FB be the 
major or longest axis, or diameter, and DC the minor 
or shorter axis or diameter, and E and K the two foci. 



These two points are obtained by taking the half of 
the major axis AB or FA, on the compasses, and. 
standing one point at D, cut the points E and K on the 
line FB, and at these points insert the pins at E and K 
as shown. Take a string as shown by the dotted lines 
and tie to the pins'at K, then stand the pencil at C 
and run the string round it and carry the string to the 
pin E, holding it tight and winding it once or twice 
around the pin, and then holding the string with the 
finger. Run the pencil around, keeping the loop of 
the string on the pencil and it will guide the latter in 
the formation of the curve as shown. When one-half 
of the ellipse is formed, the string may be used for the 
other half, commencing the curve at F or B, as the 
case may be. This is commonly called “a gardener's 
oval," because gardeners make use of it for forming 
ornamental beds for flowers, or in making curves for 






CARPENTER’S GEOMETRY 


43 


walks, etc., etc. This method of forming the curve, 
is based on the well-known property of the ellipse 
that the sum of any two lines drawn from the foci to 
their circumference 
is the same. 

Another method 
of projecting an 
ellipse is shown at 
Fig. 2, by using a 
trammel. This is an 
instrument consist¬ 
ing of two principal 
parts, the fixed part 
in the form of a cross as CD, AB, and the movable 
tracer HG. The fixed piece is made of two triangular 
bars or pieces of wood of equal thickness, joined 
together so as to be in the same plane. On one side 
of the frame when made, is a groove forming a right- 
angled cross; the groove is shown in the section at E. 
In this groove, two studs are fitted to slide easily, the 
studs having a section same as shown at F, These studs 
are to carry the tracer and guide it on proper lines. 
The tracer may have a sliding stud on the end to carry 
a lead-pencil, or it may have a number of small holes 
passed through it as shown in the cut, to carry the 
pencil. To draw an ellipse with this instrument, we 
measure off half the distance of the major axis from 
the pencil to the stud G, and half the minor axis from 
the pencil point to the stud H, then swing the tracer 
round, and the pencil will describe the ellipse required. 
The studs have little projections on their tops, that fit 
easily into the holes in the tracer, but this may be 
done away with, and two brad awls or pins may be 
thrust through the tracer and into the studs, and then 









44 


MODERN CARPENTRY 



proceed with the work. With this instrument an 
ellipse may easily be described. 

Another method, based on the trammel principle, is 
shown at Figs. 3 and 4, where the steel square is substi¬ 
tuted for the instru¬ 
ment shown in Fig, 2. 
Draw the line AB, 
bisecting it at right 
angles, draw CD. 
Set off these lines 
the required dimen¬ 
sions of the ellipse to 
be drawn. Place 
an ordinary square as 
shown. Lay the straightedge lengthwise of the figure, 
as shown in Fig. 3, and putting a pin at E against the 
square, place the pencil at F, at a point corresponding 
with the one of the figure. Next place the straight¬ 
edge, as shown in 
P'ig. 4, crosswise 
of the figure, and 
bring the pencil F 
to a point cor¬ 
responding to one 
side of the figure, 
and set a pin at G. 

By keeping the 
two pins E and G 
against the square, 
and moving the straightedge so as to carry the pencil 
from side to side, one-quarter of the figure will be 
struck. By placing the square in the same relative 
position in each of the other three-quarters, the other 
parts may be struck. 













CARPENTER’S GEOMETRY 


45 



> 8 7 <Tb 4 3 a • g 


A method,—and one that is very useful for many 
purposes,—of drawing an ellipse approximately, is 
shown in Fig. 5. It is convenient and maybe applied 
to hundreds of purposes, some of which will be illus¬ 
trated as we proceed. 

To apply this method, 
work as follows: First 
lay off the length of 
the required figure, as 
shown by AB, Fig. 5, 
and the width as shown 
by CD. Construct a 
parallelogram that shall 
have its sides tangent 
to the figure at the points of its length and width, all 
as shown by EFGH. Subdivide one-half of the end 
of the parallelogram into any convenient number of 
equal parts, as shown at AE, and one-half of its side 
in the same manner, as shown by ED. Connect these 
two sets of points by intersecting lines in the manner 
shown in the engraving. Repeat the operation for 
each of the other corners of the parallelogram. A 
line traced through the inner set of intersections will 
be a very close approximation to an ellipse. 

There are a number of ways of describing figures that 
approximate ellipses by using the compasses, some of 
them being a near approach to a true ellipse, and it is 
well that the workman should acquaint himself with 
the methods of their construction. . It is only neces¬ 
sary that a few examples be given in this work, as a 
knowledge of these shown will lead the way to the 
construction of others when required. The method 
exhibited in Fig. 6 is, perhaps, the most useful of any 
employed by workmen, than all other methods com^ 










46 


MODERN CARPENTRY 


bined. To describe it, lay off the length CD, and at 
right angles to it and bisecting it lay off the width AB. 
On the larger diameter lay off a space equal to the 
shorter diameter or width, as shown by DE. Divide 

the remainder of the 
length or larger diameter 
EC into three equal parts; 
with two of these parts 
as a radius, and R as a 
center, strike the circle 
GSFT. Then, with F as 
a center and FG as radius, 
and G as center and GF 
as radius, strike the arcs as 
shown, intersecting each other and cutting the line 
drawn through the shorter diameter at O and P respec¬ 
tively From O, through the points G and F, draw 
OL and OM, and likewise from P through the same 
points draw PK and PN. With O as center and OA 
as radius, strike the 
arc LM, and with P 
as center and with 
like radius, or PB 
which is the same, 
strike the arc KN. 

With F and G as 
centers, and with FD 
and CG which are 
the same, for radii, 
strike the arcs NM and KL respectively, thus com¬ 
pleting the figure. Another method in which the centers 
for the longer arc are outside the curve lines, is shown 
at Fig 7. Let AB be the length and CD the breadth; 
join BD through the center of the line EB, and at 










CARPENTER’S GEOMETRY 


47 


right angles to BD draw the line CF indefinitely; then 
at the points of intersection of the dotted lines will be 
found the points to describe the required ellipse. 
A method of describing 
an ellipse by the intersec¬ 
tion of lines is shown at 
Fig. 8, and which may be 
applied to any kind of an 
ellipse with longer or 
shorter axis. Let WX be 
the given major axis, and 
YA the minor axis drawn at right angles to and at the 
center of each other. 

Through Y parallel to WX draw ZT, parallel to AY, 
draw WZ and XT; divide WZ and XT into any number 
of equal parts, say four, and draw lines from the points 



of division OOO, etc., to Y. Divide WS and XS each 
into the same number of equal parts as WZ and XT, 
and draw lines from A through these last points of 
division intersecting the lines drawn from OOO, 
etc., and at these intersections trace the semi-ellipse 
WYX. The other half of the ellipse may be described 
in the same manner. 










48 


MODERN CARPENTRY 


To describe an ellipse from given diameters, by 
intersection of lines, even though the figure be on a 
rake: Let SN and QP, Fig. 9, be the given diameters, 
drawn through the centers of each other at any 
required angle. Draw QV and PT parallel to SN, 
through S draw TV parallel to QP. Divide into any 
number of equal parts PT, QV, PO, and OQ; then 
proceed as in P"ig. 8, and the work is complete 

An ellipse may be described by the intersection of 
arcs as at Fig. 10. Lay off HG and JK as the given 
axes; then find the foci as described in Fig. 1. Between 
L and L and the center M mark any number of points 
at pleasure as 1, 2, 3, 4. Upon L and L with Hi for 
radius describe arcs at O, O, O, O; upon L and with Ci 
for radius describe intersecting arcs at O, O, O, and 



O; then these points of intersection will be in the 
curve of the ellipse. The other points V, S, C, are 
found in the same manner, as follows: For the point 
V take H2 for one radius, and G2 for the other; S is 
found by taking H3 for one radius, and G3 for the 
other; C is found in like manner, with H4 for one 
radius, and G4 for the last radius, using the foci for 
centers as at first. Trace a curve through the points 
H, O, V, S, C, K, etc., to complete the ellipse. 

It frequently happens that the carpenter has to make 







CARPENTER’S GEOMETRY 


49 


the radial lines for the masons to get their arches in 
proper form, as well as making the centers for the 
same, and, as the obtaining of such lines for elliptical 
work is very tedious, I illustrate a device that may 
be employed that will obviate a great deal of labor in 
producing such lines. The instrument and the 
method of using it is exhibited at Fig. 11 and marked 
Ee. The semi-ellipse HI, or xx, may be described 
with a string or strings, the outer line being described 
by use of a string fastened to the foci F and D, with 
the extreme point at E; and the inner line, with the 
string being fastened at A and B, with the pencil point 
in the tightened string at O. The sectional line LKJ 
shows the center of the arch, and the lines SSS are at 



right angles with this vertical line. The usual method 
of finding the normal by geometry is shown at GABC, 
but the more practical method of finding it is by the 
use of the instrument, where Ee shows the normal. 
I believe the device is of French origin, and I give a 
translation of a description and use of the instrument: 
“It is made of four pieces of lath or metal put together 
so as to form a perfect rectangle and having its joints 
loose, as shown in the diagram. Considering that the 
most perfect elliptical curve is that described by a 
string from the foci (foyer) of the ellipse, draw the 
profiles of the extrados and intrados, as shown in 
Fig. 11, where your joints are to be, then take your. 






















5 ° 


MODERN CARPENTRY 


string, draw it to the point marked as at E, adjust two 
sides of your instrument to correspond with the lines 
of the string, then, from the point marked, draw a 

line passing 
through the two 
angles, E and e, 
and the line Ee 
will be the nor¬ 
mal or the radial 
line sought.” 

The oval is 
not an ellipse, 
nor are any of 
the figures ob¬ 
tained by using 
the compasses, 
as no part of an 
ellipse is a cir¬ 
cle, though it 
may approach closely to it. The oval may sometimes 
be useful to the carpenter, and it may be well to illus¬ 
trate one or two methods by which these figures may 
be described. 

Let us describe a diamond or lozenge-shaped figure, 
such as shown at Fig. 12, and then trace a curve inside 
of it as shown, touching the four sides of the figure, 
and a beautiful egg-shaped curve will be formed. For 
effect we may elongate the lozenge or shorten it at 
will, placing the short diameter at any point. This 
form of oval is much used by turners and lathe men 
generally, in the formation of pillars, balusters, newel- 
posts and turned ornamental work generally. 

An egg-shaped oval may also be inscribed in a figure 
having two unequal but parallel sides, both of which 




% 








CARPENTER’S GEOMETRY 


are bisected by the same line, perpendicular to both 
as shown in Fig. 13. These few examples are quite 
sufficient to satisfy the requirements of the workman, 
as they give the key by which he may construct any 
oval he may ever be called upon to form. 

I have dwelt rather lengthily on the subject of the 
ellipse because of its being rather difficult for the 
workman to deal with, and it is meet he should 
acquire a fair knowledge of the methods of construct¬ 
ing it. It is not my province 
to enter into all the details 
of the properties of this very 
intersecting figure, as the 
workman can find many of 
these in any good work on 
mensuration, if he should re¬ 
quire more. I may say here, 
however, that geometricians 
so far have failed to discover 
any scientific method of forming parallel ellipses, so 
that while the inside or outside lines of an ellipse can 
be obtained by any of the methods I have given, the 
parallel line must be obtained either by gauging the 
width of the material or space required, or must be 
obtained by “pricking off” with compasses or other 
aid. I thought it best to mention this as many a 
young man has spent hours in trying to solve the 
unsolvable problem when using the pins, pencil and 
string. 

There are a number of other curves the carpenter 
will sometimes meet in daily work, chief among these 
being the scroll or spiral, so .it will be well for him to 
have some little knowledge of its structure. A true 
spiral can be drawn by unwinding a piece of string that 






MODERN CARPENTRY 


has been wrapped around a cone, and this is probably 
the method adopted by the ancients in the formation 
of the beautiful Ionic spirals they produced. A spiral 

drawn by this 
method is 
shown at Fig. 
14. This was 
formed by using 
two lead-pencils 
which had been 
sharpened by 
one of those 
patent sharpen¬ 
ers and which 
gave them the 
shape seen in 
Fig. 15. A 
piece of string 
was then tied 
tightly around the pencil, and one end was wound 
round the conical end, so as to lie in notches made in 
one of the pencils; the point of a 
second pencil was pierced through the 
string at a convenient point near the 
first pencil, completing the arrange¬ 
ment shown in Fig. 15. To draw the 
spiral the pencils must be kept vertical, 
the point of the first being held firmly 
in the hole of the spiral, and the 
second pencil must then be carried 
around the first, the distance between 
the two increasing regularly, of course, as the string 
unwinds. 

This is a rough-and-ready apparatus, but a true 














CARPENTER’S GEOMETRY 


spiral can be described by it in a very few minutes. 
By means of a larger cone, spirals of any size can, of 
course, be drawn, and that portion of the spiral can be 
used which conforms to 
the required height. 

Another similar 
method is shown in 
Fig. 16, only in this 
case the string unwinds 
from a spool on a fixed 
center A, D, B. Make 
loop E in the end of 
the thread, in which 
place a pencil as shown. 

Hold the spool firmly 
and move the pencil 
around it, unwinding 
the thread. A curve will be described, as shown in 
the lines. It is evident that the proportions of the 
figure are determined by the size of the spool. Hence 

a larger or 
smaller spool 
is to be used, 
as circum¬ 
stances require. 

A simple 
method of 
forming a figure 
that corre¬ 
sponds to the 
spiral somewhat, is shown in Fig. 17. This is drawn 
from two centers only, a and e, and if the distance 
between these centers is not tbo great, a fairly smooth 
appearance will be given to the figure. The method 






54 


MODERN CARPENTRY 


of describing is simple. Take ai as radius and 
describe a semi-circle; then take ei and describe 
semi-circle 12 on the lower side of the line AB. Then 
with a2 as radius describe semi-circle above the line; 
again, with e3 as radius, describe semi-circle below 
the line AB; lastly with a3 as radius describe semi¬ 
circle above the line. 

In the spiral shown at Fig. 18 we have one drawn in 
a scientific manner, and which can be formed to 

dimensions. T o 
draw it, proceed 
as follows: Let 
BA be the given 
breadth, and the 
number of revolu¬ 
tions, say one and 
three-fourths; now 
multiply one and 
three - fourths by 
four, which equals 
seven; to which 
add three, the 
number of times a 
side of a square is 
contained in the 
diameter of the 
eye, making ten in 
all. Now divide AB into ten equal parts and set one 
from A to D, making eleven parts. Divide DB into 
two equal parts at O, then OB will be the radius of the 
first quarter OF, FE; make the side of the square, as 
shown at GF, equal to one of the eleven parts, and 
divide the number of parts obtained by multiplying 
the revolutions by four, which is seven; make the 











CARPENTER’S GEOMETRY 


55 


diameter of the eye, 12, equal to three of the eleven 
parts. With F as a center and E as a radius make the 
quarter EO; then, with G as a center, and GO as a 
radius, mark the quar¬ 
ter OJ. Take the next 
center at H and HJL 
in the quarter; so keep 
on for centers, drop¬ 
ping one part each 
time as shown by the 
dotted angles. Let 
EK be any width de¬ 
sired, and carry it 
around on the same 
centers. 

Another method of 
obtaining a spiral by 
arcs of circles is shown 
at Fig. 19, which may 
be confined to given dimensions. Proceed as follows: 
Draw SM and LK at right angles;, at the intersection 
of these lines bisect the angles by the lines NO and 
QP; and on NO and QP from the intersection each 
way set off three equal parts as shown. On 1 as center 
and iH as radius, describe the arc HK, on 2 describe 
the arc KM, on 3 describe the arc ML, on 4 describe 
the arc LR. The fifth center to describe the arc 
RT is under 1 on the line QP; and so proceed to 
complete the curve. 

There are a few other curves that may occasionally 
prove useful to the workman, and I submit an example 
or two of each in order that, should occasion arise 
where such a curve or curves are required, they may be 
met with a certain amount of knowledge of the subject. 






56 


MODERN CARPENTRY 



The first is the parabola, a curve sometimes used in 
bridge work or similar construction. Two examples 
of the curve are shown at Fig. 20, and the methods of 

describing them. 
The upper one is 
drawn as follows: 

i. Draw C8 per¬ 
pendicular to AB, 
and make it equal 
to AD. 

Next, join A8 
and B8, and divide 
both lines into the 
same number of equal parts, say 8; number them as in 
the figure; draw I, 1-2, 2-3, 3, etc., then these lines 
will be tangents to the curve; trace the curve to touch 
the center of each of those lines between the points of 
intersection. 

The lower example is described thus: I. Divide 
AD and BE, into any number of equal parts; CD and 
CE into a similar number. 

2. Draw 1, 1-2, 2, etc., parallel to AD, and from the 
points of division in AD and BE, draw lines to C. 
The points of intersection of the respective lines are 
points in the curve. 

The curves found, as in these figures, are quicker at 
the crown than a true circular segment; but, where the ' 
rise of the arch is not more than one-tenth of the 
span, the variation cannot be perceived. 

A raking example of this curve is shown in Fig. 21, 
and the method of describing it: Let AC be the ordi¬ 
nate or vertical line, and DB the axis, and B its vertex; 
produce the axis to E, and make BE equal to DB; join 
EC, EA, and divide them each into the same number 



















CARPENTER’S GEOMETRY 


57 


of equal parts, and number the divisions as shown on 
the figures. Join the corresponding divisions by the 
lines II, 22, etc., and their intersections will produce 
the contour of 
the curve. 

The h y p e r- 
bola is some¬ 
what similar in 
appearance t o 
the parabola but 
it has properties 
peculiar to it¬ 
self. It is a 
figure not much 
used in carpen- 4 
try, but it may 
be well to refer to it briefly: Suppose there be two 
right equal cones, Fig. 22, hav¬ 
ing the same axis, and cut by a 
plane Mm, Nm, parallel to that 
axis, the sections MAN, mna, 
which result, are hyperbolas. In 
place of two cones opposite to 
each other, geometricians some¬ 
times suppose four cones, which 
join on the lines EH, GB, Fig. 
23, and of which axis form two 
right lines, Ff, FT, crossing the 
center C in the same plane. 

To describe a cycloid: The 
cycloid is the curve described by 
a point in the circumference of a 
circle rolling on a straight line, 
and is described as follows: 













MODERN CARPENTRY 


58 


1. Let GH, Fig. 24, be the edge of a straight ruler, 
and C the center of the generating circle. 

2. Through C draw the diameter AB perpendicular 
to GH, and EF parallel to 
GH; then AB is the height 
of the curve, and EF is the 
place of the center of the 
generating circle at every 
point of its progress. 

3. Divide the semi-cir¬ 
cumference from B to A 
into any number of equal 
parts, say 8, and from A 
draw chords to the points 
of division. 

4. From C, with a space 
in the dividers equal to one of the divisions on the 
circle, step off on each side the same number of spaces 
as the semi-circumference is divided into, and through 
the points draw perpendiculars to GH; number them 
as in the diagram. 

5. From the points of division in EF with the 




tfig.24-. 


radius of the generating circle, describe indefinite arcs 
as shown by the dotted lines. 

6. Take the chord Ai in the dividers, and with the 
foot at I and I on the line GH, cut the indefinite arcs 









CARPENTER’S GEOMETRY 


59 


described from I and I respectively at D and D', then 
D and D' are points in the curve. 

7, With the chord A2, from 2 and 2 in GH, cut the 
indefinite arcs in J and J', with the chord A3, from 3 
and 3, cut the arcs in K and K' and apply the other 
chords in the same manner, cutting the arcs in LM, 
etc. 

8. Through the points so found trace the curve. 


B 



Each of the indefinite arcs in the diagram represents 
the circle at that point of its revolution, and the points 
DJ,K, etc., the position of the generating point B at 
each place. This curve is frequently used for the 
arches of bridges, its proportions are always constant, 
viz.: the span is equal to the circumference of the 
generating circle and the rise equal to the diameter. 
Cycloidal arches are frequently constructed which are 





6o 


MODERN CARPENTRY 


not true cycloids, but approach that curve in a greater 
or less degree. 

The epicycloidal curve is formed by the revolution 
of a circle round a circle, either within or without its 
circumference, and described by a point B, Fig. 25, in 
the circumference of the revolving circle, and Q of the 
stationary circle. 

The method of finding the points in the curve is here 
given: 

1. Draw the diameter 8, 8 and from Q the center, 
draw QB at right angles to 8, 8. 

2. With the distance QP from Q, describe an arc O, 
O representing the position of the center P throughout 
its entire progress. 

3. Divide the semi-circle BD and the quadrants D8 
into the same number of equal parts, draw chords 
from D to 1, 2, 3, etc., and from Q draw lines through 
the divisions in D8 to intersect the curve OO in 1, 
2, 3, etc. 

4. With the radius of P from I, 2, 3, etc., in* OO, 
describe indefinite arcs; apply the chords Dl, D2, etc. 
from 1, 2, 3, etc., in the circumference of Q, cutting 
the indefinite arcs in A,C,E,F, etc., which are points 
in the curve. 

We are now in a position to undertake actual work, 
and in the next chapter, I will endeavor to apply a part 
of what has preceded to practical examples, such as 
are required for every-day use. Enough geometry has 
been given to enable the workman, when he has mas¬ 
tered it all, to lay out any geometrical figure he may be 
called upon to execute; and with, perhaps, the excep¬ 
tion of circular and elliptical stairs and hand-railings, 
which require a separate study, by what has been for¬ 
mulated and what will follow, he should be able to exe¬ 
cute almost any work in a scientific manner, that may 
be placed under his control. 


PART II 


PRACTICAL EXAMPLES 
CHAPTER I 

We are now in a position to undertake the solution 
of practical examples, and I will commence this 
department by offering a few practical solutions that 
will bring into use some of the work already known to 
the student, if he has followed closely what has been 
presented. 

It is a part of the carpenter’s duty to lay out and 
construct all the wooden centers required by the brick¬ 
layer and mason for turning arches over openings of 
all kinds; therefore, it is essential he should know as 
much concerning arches as will enable him to attack 
the problems with intelligence. I have said some¬ 
thing of arches, in Part I, but not sufficient to satisfy 
all the needs of the carpenter, so I supplement with 
the following on the same subject: Arches used in 
building are named according to their curves,—cir¬ 
cular, elliptic, cycloid, parabolic, hyperbolic, etc. 
Arches are also known as three or four centered arches. 
Pointed arches are called lancet, equilateral and 
depressed. Voussoirs is the name given to the stones 
forming the arch; the central stone is called the key¬ 
stone. The highest point in an arch is called the 
crown, the lowest the springing line, and the spaces 
between the crown and springing line on either side, 
the haunches or flanks. The under, or concave, sur- 

61 


62 


MODERN CARPENTRY 


face of an arch is called the intrados or soffit, the 
upper or convex surface is called the extrados. The 
span of an arch is the width of the opening. The 
supports of an arch are called abutments, piers, or 



springing walls. This applies to the centers of wood, 
as well as to brick, stone or cement. The following 
six illustrations show the manner of getting the curves, 
as well as obtaining the radiating lines, which, as a 
rule, the carpenter will be asked to prepare for the 
mason. We take them in the following order: 

Fig. 1 . A Semi-circular Arch.—RQ is the span, and 
the line RQ is the springing line; S is the center from 



which the arch is described, and to which all joints of 
the voussoirs tend. T is the keystone of the arch. 

Fig. 2 . A Segment Arch.—U is the center from 
which the arch is described, and from U radiate all 





PRACTICAL EXAMPLES 


63 


the joints of the arch stones. The bed line of the 
arch OP or MN is called by mason builders a skew- 
back. OM is the span, and VW is the height or 
versed sine of the segment arch. 

Figs. 3 and 4 . Moorish or Saracenic Arches, one of 
which is pointed. Fig. 3 is sometimes called the 
horseshoe arch. The springing 
lines DC and ZX of both arches 
are below the centers BA and Y. 

Fig. 5 . A Form of Lintol 
Called a Platband, built in this form as a substitute 
for a segment arch over the opening of doors or windows, 
generally of brick, wedge-shaped. 

Fig. 6. The Elliptic Arch.—This arch is most per¬ 
fect when described with the trammel, and in that case 


I Tig. 5 1 


K 



the joints of the arch stones are found as follows: Let 
ZZ be the foci, and B a point on the intrados where a 
joint is required; from ZZ draw lines to B, bisect the 
angle at B by a line drawn through the intersecting 
arcs D produced for the joint to F. Joints at 1 and 2 








6 4 


MODERN CARPENTRY 


are found in the same manner. The joints for the 
opposite side of the arch may be transferred as shown. 
The semi-axes of the ellipse, HG, GK, are in the same 
ratio as GE to GA. The voussoirs near the springing 



line of the arch are thus increased in size for greater 
strength. I gave a very good description of this 
latter arch in Part I, which see. 

Another series of arches, known as Gothic arches, 
are shown as follows, with all the centers of the curve 
given, so that their formation is rendered quite simple. 
The arch shown at Fig. 7 is equilateral and its out¬ 
lines have been shown before. I repeat, however, let 
AB be the given span; on A and B as centers with 
AB as radius, describe the arcs AC and BC. 

The lancet arch, Fig. 8, is drawn as follows: DE is 
the given span; bisect DE in J, make DF and EG 
equal DJ; on F as center with FE as radius describe 



the arc EH, and on G as center describe the arc DH. 
A lancet arch, not so acute as the previous one, is 




PRACTICAL EXAMPLES 


65 


shown at Fig. 9. Let KL be the given span; bisect 
KL in M, make MP at right angles to KL and of the 
required height; connect LP, bisect LP by a line 
through the arcs R, Q produced to N; make MO equal 
MN; with N and O as centers, with NL for radius 
describe the arcs KP and LP. Fig. 10 shows a low 
or drop arch, and is obtained as follows: Let ST be 
the given span, bisect ST in W; let WX be the 
required height at right angles to TS; connect TX, 



bisect TX by a line through the arcs YZ produced to 
V, make TU equal SV; on V and U as centers with 
VT as radius describe the arcs TX and SX. Another 
Gothic arch with a still less height is shown at Fig. 
11. Suppose AB to be the given span; then divide 
AB into four equal parts; make AF and BG equal 
AB, connect FE and produce to D; with CA as radius, 
on C and E, describe the arcs AD and BK; on F and 
G as centers, describe the arcs JK and DK. 

Another four-centered arch of less height is shown 
at Fig. 12. Let SI be the given span, divide into six 
equal parts; on R and Q as centers with RQ as radius 
describe the arcs QV and RV, connect QV and RV and 
produce to L and M; on R and Q as centers with QT as 






66 


MODERN CARPENTRY 


radius describe the arcs TP and SO; on L and M as 
centers describe the arcs PN and ON. 

To describe an equilateral Ogee arch, like Fig. 13, 
proceed as follows: Make YZ the given span; make 



YX equal YZ, bisect YZ in A; on A as center with 
AY as radius describe the arcs YB and ZC; on B and 
X as centers describe the arcs BD and XD, and on G 
and X as centers describe the arcs CE and XE, on E 
and D as centers describe the arcs BX and CX. 

Fig. 14 shows the method of obtaining the lines for 
an Ogee arch, having a height equal to half the span. 
Suppose FH to be the span, divide into four equal 
parts, and at each of the points of division draw lines 
LN, KG and JO at right angles to FH; with LF for 
radius on L and J describe the quarter circles FM and 
HP; and with the same radius on O and N describe 
the quarter circles PG and MG. 

These examples—all or any of them—can be made use 
of in a great number of instances. Half of the Ogee 
curve is often employed for veranda rafters, as for the 
roofs of bay-windows, for tower roofs and for bell 
bases, for oriel and bay-windows, and many other 
pieces of work the carpenter will be confronted with 
from time to time. They also have value as aids in 
forming mouldings and other ornamental work, as for 


PRACTICAL EXAMPLES 


67 


example Fig. 15, which shows a moulding for a base 
or other like purpose. It is described as follows: 
Draw AB; divide it into five 
equal parts; make CD equal to 
four of these. Through D draw 
DF parallel with AB. From D, 
with DC as radius, draw the arc 
CE. Make EF equal to DE; di¬ 
vide EF into five parts; make the 
line above F equal to one of these; 
draw FG equal to six of these. 

From G, with radius DE, describe 
the arc; bisect GF, and lay the 
distance to H. It is the center of 
the curve, meeting the semi-circle 
described from M. Join NO, OS, 
and the moulding is complete. 

The two illustrations shown at 
Figs. 16 and 17 will give the stu¬ 
dent an idea of the manner in 
which he can apply the knowledge he has now obtained, 
and it may not be out of place to say that with a little 
ingenuity he can form almost any sort of an ornament 
he wishes by using this knowledge. The two illustra¬ 
tions require no explanation as their formation is self- 
evident. Newel posts, balusters, pedestals and other 
turned or wrought ornaments, may be designed easily 
if a little thought be brought to bear on the subject. 

The steel square is a great aid in working out prob¬ 
lems in carpentry, and I will endeavor to show, as 
briefly as possible, how the square can be applied to 
some difficult problems, and insure correct solutions. 

It is unnecessary to give a full and complete descrip¬ 
tion of the steel square. Every carpenter and joiner is 








48 


MODERN CARPENTRY 


supposed to be the possessor of one of these useful 
tools, and to have some knowledge of using it. It is 
not everyone, however, who thoroughly understands 
its powers or knows how to employ it in solving all 



the difficulties of framing, or to take advantage of its 
capabilities in laying out work. While it is not my 
intention to go deeply into this subject in this vol¬ 
ume, as that would lengthen it out to unreasonable 
limits, so it must be left for a separate work, yet there 
are some simple things connected with the steel square, 
that I think every carpenter and joiner should know, 
no matter whether he intends to go deeper into the 
study of the steel square or not. One of these things 
is the learning to read the tool. Strange as it may 





































PRACTICAL EXAMPLES 


69 


appear, not over one in fifty of those who use the 
square are able to read it, or in other words, able to 
explain the meaning and uses of the figures stamped 
on its two sides. The following,will assist the young 
fellows who want to master the subject. 

The square consists of two arms, at right angles to 
each other, one of which is called the blade and which 
is two feet long, and generally two inches wide. The 
other arm is called the tongue, and may be any 
length from twelve to eighteen inches, and 1% to 
2 inches in width. The best square has always a 
blade 2 inches wide. Squares made by firms of repute 
are generally perfect and require no adjusting or 
“squaring.’* 

The lines and figures formed on squares of different 
make sometimes vary, both as to their position on the 
square and their mode of application, but a thorough 
understanding of the application of the scales and 
lines shown on any first-class tool, will enable the stu¬ 
dent to comprehend the use of the lines and figures 
exhibited on any good square. 

It is supposed the reader understands the ordinary 
divisions and subdivisions of the foot and inch into 
twelfths, inches, halves, quarters, eighths and six¬ 
teenths, and that he also understands how to use that 
part of the square that is subdivided into twelfths of 
an inch. This being conceded, we now proceed to 
describe the various rules as shown on all good squares. 
Sometimes the inch is subdivided into thirty-seconds, 
in which the subdivision is very fine, but this scale 
will be found very convenient in the measure¬ 
ment of drawings which are made to a scale of 
half, quarter, one-eighth or one-sixteenth of an inch 
to a foot. 


70 


MODERN CARPENTRY 



intended for taking 


In the illustration Fig. 
18, will be noticed a series 
of lines extending from 
the junction of the blade 
and tongue to the four- 
inch limit. From the 
figures 2 to 3 these lines 
are crossed by diagonal 
lines. This figure, reach¬ 
ing from 2 to 4, is called 
a diagonal scale, and is 
hundredths of an inch The 


pmr[x 


[TTTS 

T7JTT 

TTTTT 

'4 



ISSl 



pspnpppT ii|u|ii|H|ii|)i|H|in 




iLlLlll 


ill 


liiidi 


Fig. 19 

lengths of the lines between the 
diagonal and the perpendicular 
are marked on the latter. Primary 
divisions are tenths, and the junc¬ 
tion of the diagonal lines with the 
longitudinal parallel lines enables 
the operator to obtain divisions of 
one-hundredth part of an inch; as 
for example, if we wish to obtain 
twenty-four hundredths we operate 
on the seventh line, taking five 
primaries and the fraction of the 
sixth where the diagonal inter¬ 
sects the parallel line, as shown 














































































PRACTICAL EXAMPLES 


7i 


by the “dots’* on the compasses, and this gives us the 
distance required. 

The use of the scale is obvious, and needs no furtner 
explanation, as the dots or points are shown. 

The lines of figures running across the blade of the 
square, as shown in Fig. 19, forms what is a very con¬ 
venient rule for determining the amount of material in 
length or width of stuff. To use it proceed as fol¬ 
lows: If we examine we will find under the figure 12, 
on the outer edge of the blade, where the length of the 
boards, plank or scantling to be measured is given, 
and the answer in feet and inches is found under the 
inches in width that the board, etc., measures. For 
example, take a board nine feet long and five inches 
wide, then under the figure 12, on the second line, 
will be found the figure 9, which is the length of the 
board; then run along this line to the figure directly 
under the five inches (the width of the board) and we 
find three feet nine inches, which is the correct answer 
in ‘ board measure.” If the stuff is three inches thick 
it is trebled, etc., etc. If the stuff is longer than any 
figures shown on the square it can be measured as 
above and doubling the result. This rule is calcu¬ 
lated, as its name indicates, for board measure, or for 
surfaces 1 inch in thickness. It may be advantageously 
used, however, upon timber by multiplying the result 
of the face measure of one side of a piece by its depth 
in inches. To illustrate, suppose it be required to 
measure a piece 25 feet long, 10x14 inches in size. 
For the length we will take 12 and 13 feet. For the 
width we will take 10 inches, and multiply the result 
by 14. By the rule a board 12 feet long and 10 inches 
wide contains 10 feet, and one 13 feet long and 10 
inches wide, 10 feet 10 inches. Therefore, a board 25 
feet long and 10 inches wide must contain 20 feet and 


MODERN CARPENTRY 


12 

10 inches. In the timber above described, however, 
we have what is equivalent to 14 such boards, and 
therefore we multiply this result by 14, which gives 
291 feet and 8 inches the board measure. 

Along the tongue of the square following the diag¬ 
onal scale is the brace rule, which is a very simple and 
very convenient method of determining the length of 
any brace of regular run. The length of any brace 
simply represents the hypothenuse of a right-angled 
triangle. To find the hypothenuse extract the square 
root of the sum of the squares of the perpendicular 
and horizontal runs. For instance, if 6 feet is the 
horizontal run and 8 feet the perpendicular, 6 squared 
equals 36, 8 squared equals 64; 36 plus 64 equals 100, 
the square root of which is 10. These are the rules 
generally used for squaring the frame of a building. 

If the run is 42 inches. 42 squared is 1764, double 
that amount, both sides being equal, gives 3528, the 
square root of which is, in feet and inches, 4 feet 11.40 
inches. 

In cutting braces always allow in length from a six¬ 
teenth to an eighth of an inch more than the exact 
measurement calls for. 

Directly under the half-inch marks on the outer edge 
of the back of the tongue, Fig. 19, will be noticed two 
figures, one above the other. These represent the run 
of the brace, or the length of two sides of a right- 
angled triangle; the figures immediately to the right 
represent the length of the brace or the hypothenuse. 
For instance, the figures f 5, and 80.61 show that the run 
on the post and beam is 57 inches, and the length of the 
brace is 80.61 inches. 

Upon some squares will be found brace measure¬ 
ments given, where the run is not equal, as lf.30. It 
toil! be noticed that the last set of figures are each just 


PRACTICAL EXAMPLES 


73 


three times those mentioned in the set that are usually 
used in squaring a building. So if the student or 
mechanic will fix 
in his mind the 
measurements of a 
few runs, wich the 
length of braces, 
he can readily 
work almost any 
length required. 

Take a run, for 
instance, of 9 
inches on the 
beam and 12 
inches on the post. 

The 1 e n gt h of 
brace is 15 inches. In a run, therefore, of 12, 16, 20, ot 
any number of times above the figures, the length ov 
the brace will bear the same proportion to the run st 
the multiple used. Thus if you multiply all the fig¬ 
ures by 3 you will have 36 and 48 inches for the run, 
and 60 inches for the brace, or to remember still more 
easily, 3, 4 and 5 feet. 

There is still another and an easier method of obtain¬ 
ing the lengths of braces by aid of the square, also the 
bevels as may be seen in Fig. 20, where the run is 3 
feet, or 36 inches, as marked. The length and bevels 
of the brace are found by applying the square three 
times in the position as shown; placing 12 and 12 on 
the edge of the timber each time. By this method 
both length and bevel are obtained with the least 
amount of labor. Braces having irregular runs may 
be oberated in the same manner. For instance, sup¬ 
pose we wish to set in a brace where the run is 4 
feet and 3 feet; we simply take 9 inches on the 













MODERN CARPENTRY 


tongue and 12 inches on the blade and apply the 
square four times, as shown in 
Fig. 21, where the brace is 
given in position. Here we 
get both the proper length and 
the exact bevels. It is evident 
from this that braces, regular 
or irregular, and of any length, 
may be obtained with bevels for 
same by this method, only care 
must be taken in adopting the 
figures for the purpose. 

If we want a brace with a two- 
foot run and a four-foot run, it must be evident that 
as two is the half of four, so on the square take 12 
inches on the tongue, and 6 inches on the blade, apply 
four times and we have the length and the bevels of a 
brace for this run. 

For a three-by-four foot run take 12 inches on the 
tongue and 9 inches on the blade, and apply four 
times, because as 3 feet is ^ of four feet, so 9 inches 
is of 12 inches. 

While on the subject of braces I submit the follow¬ 
ing table for determining the length of braces for any 
run from six inches to fourteen feet. This table has 
been carefully prepared and may be depended upon as 
giving correct measurements. Where the runs are 
regular or equal the bevel will always be a miter or 
angle of 45°, providing always the angle which the 
brace is to occupy is a right angle—a “square." If 
the run is not equal, or the angle not a right angle, 
then the bevels or “cuts” will not be miters, and will 
have to be obtained either by taking figures on the 
square or by a scaled diagram. 









PRACTICAL EXAMPLES 


75 


TABLE 


Length of 

Run 

Length of 
Brace 

ft. in. 

ft. 

in. 

ft. 

in. 

6 x 


6 = 


8.48 

6 x 


9 “ 


I0.8I 

9 ^ 


9 = 

I 

O .72 

I O X 

I 

o = 

I 

4-97 

I o x 

i 

3 = 

I 

7.20 

I 3 x 

i 

3 = 

I 

9-23 

i 3 x 

I 

6 = 

I 

n. 43 

i 6 x 

i 

6 - 

2 

1-45 

I 6 x 

i 

9 = 

2 

365 

I 9 x 

j 

9 = 

2 

5.69 

I 9 x 

2 

o = 

2 

7.89 

2 0 X 

2 

o = 

2 

9-94 

2 0 X 

2 

3 = 

3 

O.I 2 

2 O X 

2 

6 = 

3 

2 . 4 I 

2 3 X 

2 

6 = 

3 

4.36 

2 6 X 

2 

6 = 

3 

6.42 

2 6 X 

2 

9 = 

3 

8.59 

2 9 X 

2 

9 = 

3 

10.66 

2 9 x 

3 

o = 

4 

0.83 

3 o X 

3 

o = 

4 

2.91 

3 o X 

3 

3 = 

4 

5.02 

3 o x 

3 

6 = 

4 

7-3i 

3 o x 

3 

9 = 

4 

9.62 

3 3 x 

3 

3 = 

4 

7.15 

3 3 x 

3 

6 = 

4 

9-3i 

3 3 x 

3 

9 = 

4 

11.54 

3 3 x 

4 

o = 

5 

1.84 

3 6 x 

3 

6 = 

4 

n.39 

3 6 x 

3 

9 - 

5 

1.55 

3 6 x 

4 

o = 

5 

378 

3 9 x 

3 

9 = 

5 

3-63 

3 9 x 

4 

o = 

5 

579 

4 o x 

4 

0 = 

5 

7.88 

4 o x 

4 

3 = 

5 

10.03 

4 o x 

4 

6 = 

6 

0.25 

4 0 x 

4 

9 = 

6 

271 

4 o x 

5 

o = 

6 

4.83 


Length of Length of 

Run Brace 


ft. in. 

ft. in. 

ft. 

in. 

4 3 x 

4 3 = 

6 

0.12 

4 3 x 

46- 

6 

2.27 

4 3 x 

4 $ = 

6 

4-49 

4 3 x 

5 0 - 

6 

6.74 

4 6 x 

4 6 = 

6 

476 

4 6 x 

4 9 = 

6 

6.51 

4 6 x 

5 0 = 

6 

8.72 

4 9 X 

4 9 = 

6 

8.61 

4 9 X 

5 0 = 

6 

10.75 

5 0 X 

5 0 - 

7 

O.85 

5 3 x 

5 3 ^ 

7 

5.09 

5 6 x 

5 6 = 

7 

9-33 

5 9 x 

5 9 = 

8 

1:58 

6 0 x 

60 = 

8 

5.82 

6 3 x 

6 3 = 

8 10.06 

6 6 x 

6 6 = 

9 

2.30 

6 9 x 

69 = 

9 

6.55 

7 0 X 

70 = 

9 

10.79 

7 3 x 

7 3 = 

10 

3.03 

7 6 x 

76 = 

10 

7.28 

7 9 x 

7 9 = 

10 

11.52 

8 0 x 

80 = 

11 

376 

8 3 x 

8 3 = 

11 

8.00 

8 6 x 

8 6 = 

12 

0.24 

8 9 x 

8 9 = 

12 

4.49 

9 0 X 

90 = 

12 

873 

9 6 x 

96 = 

13 

5.22 

10 0 X 

10 0 = 

14 

1.70 

10 6 x 

10 6 = 

14 10.19 

II 0 X 

11 0 = 

15 

6.67 

11 6 x 

116 = 

16 

3.16 

12 0 X 

12 0 = 

16 11.64 

12 6 x 

12 6 = 

7 

8.13 

13 0 X 

13 0 = 

18 

4.61 

13 6 x 

13 6 = 

19 

1.10 

[4 0 X 

14 0 = 

19 

9.58 






7 6 


MODERN CARPENTRY 


vl 1 1 

1 LI 

TIT 

TT71 


2 

3 

A 



*1 • • • 

• • • * / 




\ 


LU 

-LU. 

rml 


Fig, 22,; 


There is on the 
tongue of the square 
a scale called the 
“octagonal s c a 1 e .” 
This is generally on 
the opposite side to 
Fig. 22 exhibits a por- 
It is 


the scales shown on Fig. 19. 
tion of the tongue on which this scale is shown, 
the central division on which the number 10 
is seen along with a number of divisions. < 
It is used in this way: If you have a stick 
10 inches square which you wish to dress up * 
octagonal, make a center mark on each 
face, then with the compasses, take 10 of the N 
spaces marked by the short cross-lines in the < 
middle of the scale, and layoff this distance 
each side of the center lines, do the same at \ 
the other end of the stick, and strike a chalk 
line through these marks. Dress off the cor¬ 
ners to the lines, and the stick will be octag¬ 
onal. If the stick is not straight it must be 
gauged, and not marked with the chalk line. 
Always take a number of spaces equal to the 
square width of the octagon in inches. This 
scale can be used for large octagons by 
doubling or trebling the measurements. 

On some squares, there are other scales, 
but I do not advise the use of squares that 
are surcharged with too many scales and fig¬ 
ures, as they lead to confusion and loss of time. 

It will now be in order to offer a few 
things that can be done with the steel 
square, in a shorter time than by applying 
any other methods. If we wish to get the 























PRACTICAL EXAMPLES 


77 


length and bevels for any common rafter it can be done 
on short notice by using the square as shown in 
Fig. 23. The pitch of the roof will, of course, gov¬ 
ern the figures to be employed on the blade and tongue. 
For a quarter pitch, the figures must be 6 and 12. For 
half pitch, 12 and 12 must be used. For a steeper 
pitch, 12 and a larger figure must be used according 
to the pitch required. For the lower pitches, 8 and 
12 gives a one-third pitch and 9 and 12 a still steeper 
pitch; and from this the workman can obtain any pitch 
he requires. If the span is 24 feet, the square must be 
applied 12 times, as 12 is half of 24. And so with 
any other span: The square must be applied half as 
many times as there are feet in the width. This is 
self-evident. The bevels and lengths of hip and val¬ 
ley rafters may be obtained in a similar manner, b£ 
first taking the length of the diagonal line between 12 
and 12, on the square, which is 17 inches in round 
numbers. Use this figure on the blade, and the “rise" 
whatever that may be, on the tongue. Suppose we 
have a roof of one-third pitch, which has a span 
of 24 feet; then 8, which is one-third of 24, will be 
the height of the roof at the point or ridge, from the 
base of the roof on a line with the plates. For 
example, always use 8, which is one-third of 24, on 
tongue for altitude; 12, half the width of 24, on blade 
for base. This cuts common rafter. Next is the hip 
rafter. It must be understood that the diagonal of 12 
and 12 is 17 in framing t as before stated, and the hip 
is the diagonal of a square added to the rise of roof; 
therefore we take 8 on tongue and 17 on blade; run 
the same number of times as common rafter. To cut 
jack rafters, divide the number of openings for com¬ 
mon rafter. Suppose we have 5 jacks, with six open- 


7 8 


MODERN CARPENTRY 


ings, our common rafter 12 feet long, each jack would 
be 2 feet shorter, first 10 feet, second 8 feet, third 6 
feet, and so on. The top down cut the same as cut of 
common rafter; foot also the same. To cut miter to 
fit hip: Take half the width of building on tongue and 
length of common rafter on blade; blade gives cut. 
Now find the diagonal of 8 and 12, which is I4 T \, take 
12 on tongue, on blade; blade gives cut. The 

hip rafter must be beveled to suit; height of hip on 
tongue, length of hip on blade; tongue gives bevel. 
Then we take 8 on tongue, 2 >% on blade; tongue gives 
the bevel. Those figures will span all cuts in putting 
on cornice or sheathing. To cut bed moulds for gable 
to fit under cornice, take half width of building on 

tongue, length of 
common rafter on 
blade; blade gives 
cut; machine mould¬ 
ings will not mem¬ 
ber, but this gives a 
solid joint; and to 
member properly it 
is necessary to make moulding by hand, the diagonal 
plumb cut differences. To cut planceer to run up 
valley, take height of rafter on tongue, length of rafter 
on blade; tongue gives cut. The plumb cut takes the 
height of hip rafter on tongue, length of hip rafter on 
blade; tongue gives cut. These figures give the cuts 
tor one-third pitch only, regardless of width of build¬ 
ing. The construction of roofs generally will be taken 
up in another chapter. 

A ready way of finding the length and cuts for cross¬ 
bridging is shown at Fig. 24. If the joists are 8 inches 
wide and 16 inches centers, there will be 14 inches 











PRACTICAL EXAMPLES 


oetween. Place the square on 8 and 14, and cut on 8, 
Qnd you have it. The only point to observe is that the 
8 is on the lower side of the piece of bridging, while the 
14 is on the upper, and not both on same side of tim¬ 
ber, as in nearly all work. Bridging for any depth of 
joists, to any rea¬ 
sonable distance of 
joists apart, may be 
obtained by this 
method. A quick 
way of finding the 
joists for laying out 
timber to be worked from ^he square to an octagon sec¬ 
tion is shown at Fig. 25. Lay your square diagonally 
across your timber and mark at 7 and 17, which gives 
corner of octagon. The figures 7 and 17, on either 
a square or two-foot pocket rule, wi;~n laid on a board 
or piece of timber as shown, always define the points 
where the octagonal angle or arris should be. 

Fig. 26 '•hows a 
rapid method of 
dividing anything 
into several equal 
parts. If the board 
is io }4 inches wide, 
lay the square from 
heel to 12, and mark at 3, 6 and 9, and you have it 
divided into four equal parts. Any width of board or 
any number of parts may be worked with accuracy 
under the same method. 

A method for obtaining the “cuts” for octagon and 
hexagon joints is shown at Fig. 27. Lay off a quarter 
circle XA, with C as a center; then along the hori¬ 
zontal line AB the scuiare' is laid with 12" on the blade 




Fig. 25, 










8o 


MODERN CARPENTRY 


at the center C, from which the quadrant was struck. 
If we divide this quadrant into halves, we get the point 
E, and a line drawn from 12" on the blade of the 
square and through the point E, we cut the tongue of 
the square at 12" and through to O, and the line thus 
drawn makes an angle of 45 0 , a true miter. If we 
divide the quadrant between E and X, and then draw 
a line from C, and 12" on the blade of the square, cut¬ 
ting the dividing point D, we get the octagon cut, 
which is the line DC. Again, if we divide the space 



between E and X into three equal parts, making GC 
one of these parts, and draw a line from C to G cutting 
the tongue of the square at 7", we get a cut that will 
give us a miter for a hexagon; therefore, we see from 
this that if we set a steel square on any straight edge 
or straight line, 12" and 12" on blade and tongue on 
the line or edge, we get a true miter by marking along 
the edge of the blade. For an octagon miter, we set 
the blade on the line at 12", and the tongue at 5", and 
we get the angle on the line of the blade—nearly; and, 
for a hexagon cut, we place the blade at 12" on the 










PRACTICAL EXAMPLES 81 

line, and the tongue at 7", and the line of the blade 
gives the angle of cut—nearly. The actual figure for 
octagon is 4§J, but 5" is close enough; and for a hexa¬ 
gon cut, the exact figures are 12" and 6i|, but 12" and 
7" is as near as most workmen will require, unless the 
cut is a very long one. 

The diagram shown at Fig. 28 illustrates a method 
of defining the pitches of roofs, and also gives the fig¬ 
ures on the square for laying out the rafters for such 
pitches. By a very common usage among carpenters 
and builders, the pitch of a roof is described 
by indicating what fraction the rise is of the 
span. If, for example, the span is 24 feet 
(and here it should be remarked that the dia¬ 
gram shows only one-half the span), then 6 
feet rise would be called 
quarter pitch, because 6 is 
one-quarter of 24. The rule, 
somewhat arbitrarily ex¬ 
pressed, that is applicable 




r.J-i 1 1 1 1 l i-i f 1 1 1 1 ) ■ 1 ■ 1 1 t 


22 2 / 20 /9 


r/ /<> /s ** 4 ? /z // 



in such cases in roof framing where the roof is one- 
quarter pitch, is as follows: Use 12 of the blade, and 
6 of the tongue. For other pitches use the figures 
appropriate thereto in the same general manner. 

The diagram indicates the figures for sixth pitch, 
quarter pitch, third pitch and half pitch. The first 
three of these are in very common use, although the 
latter is somewhat exceptional. 

It will take but a moment’s reflection upon the part 













82 


MODERN CARPENTRY 


of a practical man, with this diagram before him, to 
perceive that no changes are necessary in the rule 
where the span is more or less than 24 feet. The cuts 
are the same for quarter pitch irrespective of the 
actual dimensions of the building. The square in all 
such cases is used on the basis of similar triangles. 
The broad rule is simply this: To construct with the 
square such a triangle as will proportionately and cor¬ 
rectly represent the full size, the blade becomes the 
base, the tongue the altitude or rise, while the hypoth- 


enuse that results rep¬ 
resents the rafter. The 
necessary cuts are 
shown by the tongue 
and blade respectively. 



In order to give a gen¬ 
eral idea of the use of 
the square I herewith ap¬ 
pend a few illustrations 


of its application in framing a roof of, say, one-third 
pitch, which will be supposed to consist of common 
rafters, hips, valleys, jack rafters and ridges. Let it 
be assumed that the building to be dealt with measures 
30 feet from outside to outside of wall plates; the toe 
of the rafters to be fair with the outside of the wall 
plates, the pitch being one-third (that is the roof rises 
from the top of the wall plate to the top of the ridge, 
one-third of the width of the building, or 10 feet), the 
half width of the building being 15 feet. Thus, the 
figures for working on the square are obtained; if 
other figures are used, they must bear the same relative 
proportion to each other. 

To get the required lengths of the stuff, measure 
across the corner of the square, from the 10-inch mark 






PRACTICAL EXAMPLES 


83 


on the tongue to the 15-inch mark on the blade, 
Fig. 29. This gives 18 feet as the length of the 
common rafter. To get the bottom bevel or cut to 
fit on the wall plate, lay the square flat on the side of 
the rafter. Start, say, at the right-hand end, with the 
blade of the square to the right, the point or angle of 
the square away from you, and the rafter, with its 
back (or what will be the top edge of it when it is 
fixed) towards you. Now place the 15-inch mark of 
the blade and the 10-inch mark of the tongue on the 
corner of the rafter—that is, towards you—still keeping 
the square laid 
flat, and mark 
along the side of 
the blade. This 
gives the bottom 
cut, and will fit 
the wall plate. 

Now move the 
square to the other 
end of the rafter, place it in the same position as 
before to the 18-foot mark on the rafter and to the 
io-inch mark on the tongue, and the 15-inch mark on 
the blade; then mark alongside the tongue. This 
gives ‘the top cut to fit against the ridge. To get the 
length of the hip rafter, take 15 inches on the blade 
and 15 inches on the tongue of the square, and measure 
across the corner. This gives 2i T \ inches. Now take 
this figure on the blade and 10 inches on the tongue, 
then measuring across the corner gives the length of 
the hip rafter. 

Another method is to take the 17-inch mark on the 
blade and the 8-inch mark on the tongue and begin as 
with the common rafter, as at Fig. 30. Mark along 






MODERN CARPENTRY 


*4 

the side of the blade for the bottom cut. Move the 
square to the left as many times as there are feet in 
the half of the width of the building (in the present 

case, as we have seen, 15 feet is half the width), keep¬ 

ing the above mentioned figures 17 and 8 in line with 
the top edge of the hip rafter; 
step it along just the same as 
when applying a pitch board on 
a stair-string, and after moving 
it along 15 steps, mark along¬ 
side the tongue. This gives the 
top cut or bevel and the length. 
The reason 17 and 8 are taken 
on the square is that 12 and 8 rep¬ 
resent the rise and run of the 
common rafter to I foot on plan, 
while 17 and 8 correspond with the plan of the hips. 

To get the length of the jack rafters, proceed in the 
same manner as for common or hip rafters; or alter¬ 
nately space the jacks and divide the length of the com¬ 
mon rafter into the same 
number of spaces. This 
gives the length of each 
jack rafter. 

To get the bevel of the 
top edge of the jack rafter, 

Fig. 31, take the length, 

14^4 of the common rafter 
on the blade and the run of the common rafter on the 
tongue, apply the square to the jack rafter, and mark 
along the side of the blade; this gives the bevel or cut. 
The down bevel and the bevel at the bottom end are 
the same as for the common rafter 

To get the bevel for the side of the purlin to fit 













PRACTICAL EXAMPLES 85 

against the hip rafter, place the square flat against the 
side of the purlin, with 8 inches on the tongue and 
1434 inches on the blade, Fig. 32. Mark alongside of 
the tongue. This gives the side cut or bevel. The 
14^4 inches is the length of the common rafter to the 
i-foot run, and che 8 inches represent the rise. 

For the edge bevel of purlin, lay the square flat 
against the edge of purlin with 12 inches on the tongue 
and 1456 'laches on the blade, as at Fig. 33, and mark 
along the side of the 
tongue. This gives 
the bevel or cut for the 
edge of the purlin. 

The rafter patterns 
must be cut half the 
thickness of ridge 
shorter; and half the 
thickness of the hip rafter allowed off the jack rafters. 

These examples of what may be achieved by the aid 
of the square are only a few of the hundreds that can 
be solved by an intelligent use of that wonderful instru¬ 
ment, but it is impossible in a work of this kind to 
illustrate more than are here presented. The subject 
will be dealt with at length in a separate volume. 






CHAPTER II 


GENERAL FRAMING AND ROOFING 

Heavy framing is now almost a dead science in this 
country unless it be in the far west or south, as steel 
and iron have displaced the heavy timber structures 
that thirty or forty years ago were so plentiful in 
roofs, bridges and trestle-work. As it will not be 



necessary to go deeply into heavy-timber framing, 
therefore I will confine myself more particularly to the 
framing of ballon buildings generally. 

A ballon frame consists chiefly of a frame-work of 
scantling. The scantling may be 2 x 4 inches, or any 
other size that may be determined. The scantlings are 
spiked to the sills, or are nailed to the sides of the 
joist which rests on the sills, or, as is sometimes the 
case, a rough floor may be nailed on the joists, 

86 










































































































PRACTICAL EXAMPLES 


87 


and on this, ribbon pieces 
of 2 x 4-inch stuff are 
spiked around to the outer 
edge of the foundation, 
and onto these ribbon 
pieces the scantling is 
placed and “toe-nailed” 
to them. The doors and 
windows are spaced off as 
shown in Fig. 34, which 
represents a ballon frame 
and roof in skeleton condition. These frames are 
generally boarded on both sides, always on the out¬ 
side. Sometimes the boarding on the outside is nailed 
on diagonally, but more 
frequently horizontally, 
which, in my opinion, is 
the better way, providing 
always the boarding is dry 
and the joints laid close. 

The joists are laid on 
“rolling,” that is, there 
are no gains or tenons em¬ 
ployed, unless in trimmers 
or similar work. The 
joists are simply “toe- 
nailed” onto sill plates, or 
ribbon pieces, as shown in the illustration. Sometimes 
the joists are made to rest on the sills, as shown in 
Fig. 35, the sill being no more 
than a 2 x 4-inch scantling laid 
in mortar on the foundation, the 
outside joists forming a sill for 
the £ide studs. A better plan is 







































88 


MODERN CARPENTRY 


shown in Fig. 36, which gives a method known as a 
“box-sill.” The manner of construction is very 
simple. 



All joists in a building of this kind must be bridged 
similar to the manner shown in Fig. 37, about every 
eight feet of their length; in spans less than sixteen 
feet, and more than eight feet, a row of bridging 
should always be put in midway in the span. Bridg¬ 
ing should not be less than 
I to 1 y 2 inches in section. 

In trimming around a 
chimney or a stair well-hole, 
several methods are em¬ 
ployed. Sometimes the 
headers and trimmers are 
made from material twice as 
thick and the same depth as 
the ordinary joists, and the intermediate joists are 
tenoned into the header, as shown in Fig. 38. Here 
we have T, T, for header, and T, J, T, J, for trimmers, 
and by /, for the ordinary joists. In the western, and 
also some of the central States, the trimmers and 
headers are made up of two thicknesses, the header 
being mortised to secure the ends of the joists. The 
























































































PRACTICAL EXAMPLES 


89 


two thicknesses are 
well nailed together. 

This method is exhib¬ 
ited at Fig. 39., which 
also shows one way to 
trim around a hearth; 

C shows the header 
with trimmer joists 
with tusk tenons, keyed 
solid in place. 

Frequently it hap¬ 
pens that a chimney 
rises in a building from 
its own foundation, disconnected 
from the walls, in which case the 
chimney shaft will require to be 
trimmed all around, as shown in 





4 Fig. 41 # 


Fig. 40. In cases of 
this kind the trim¬ 
mers A, A, should be 
made of stuff very 
much thicker than 
the joists, as they 
have to bear a double 
burden; B, B shows 
the heading, and C, 
C, C, C the tail joists. 
B, B, should have a 
thickness double that 
of C, C, etc., and A, 
A should at least be 


















































































9 o 


MODERN CARPENTRY 


three times as stout as C, C. This will to some extent 
equalize the strength of the whole floor, which is a 
matter to be considered in laying down floor timbers, 
for a floor is no stronger than its weakest part. 

There are a number of devices for trimming around 
stairs, fire-places and chimney-stacks by which the 
cutting or mortising of the timbers is avoided. One 
method is to cut the timbers the exact length, square 


in the ends, and then insert 
iron dowels—two or more— 
in the ends of the joists, 
and then bore holes in the 
trimmers and headers to suit, 
and drive the whole solid 
together. The dowels are 
made from ^-inch or i-inch 
round iron. Another and a 
better device is the “bridle 
iron,” which may be hooked 
over the trimmer or header, 
as the case may be, the stir¬ 



rup carrying the abutting timber, as shown in Fig. 41. 
These bridle irons” are made of wrought iron— 
2 x 2^4 inches, or larger dimensions if the work requires 
such; for ordinary jobs, however, the size given will 
be found plenty heavy for carrying the tail joists, and a 
little heavier may be employed to carry the header. 
This style of connecting the trimmings does not hold 
the frame-work together, and in places where there is 
any tendency to thrust the work apart, some provision 
must be made to prevent the work from spreading. 

In trimming for a chimney in a roof, the “headers,” 
“stretchers” or “trimmers,” and “tail rafters,” may 
be simply nailed in place, as there is no great weight 








PRACTICAL EXAMPLES 


beyond snow and wind pressure to carry, therefore 
the same precautions for strength are not necessary. 
The sketch shown at Fig. 42 explains how the chimney 
openings in the roof may be trimmed, the parts being 
only spiked together. A shows a hip rafter against 
which the cripples on both sides are spiked. The 
chimney-stack is shown in the center of the roof — 
isolated—trimmed on the four sides. The sketch is 



*elf-explanatory in a measure, and should be easily 
understood. 

An example or two showing how the rafters may be 
connected with the plates at the eaves and finished for 
cornice and gutters, may not be out of place. A sim¬ 
ple method is shown at Fig. 43, where the cornice is 
complete and consists of a few members only. The 
gutter is attached to the crown moulding, as shown. 

Another method is shown at Fig. 44, this one 
being intended for a brick wall having mailing courses 
over cornice. The gutted is built in of wood, and i* 
















92 


MODERN CARPENTRY 


lined throughout with galvanized iron This makes a 
substantial job and may be used to good purpose on 
brick or stone warehouses, factories or similar build¬ 
ings. 

Another style of rafter finish is 
shown at Fig. 45, which also shows 
scheme of cornice, A similar fin¬ 
ish is shown at Fig. 46, the cor¬ 
nice being a little differ¬ 
ent. In both these exam¬ 
ples, the gutters are of 
wood, which should be 
lined with sheet metal of 
some sort in order to pre¬ 
vent their too rapid de¬ 
cay. At Fig. 47 a rafter 
finish is shown which is 
intended for a veranda or porch. 

Here the construction is very simple. 

The rafters are dressed and cut on 
projecting end to represent brackets 
and form a finish 

From these examples the workman will get sufficient 
ideas for working his rafters to suit almost any condi¬ 
tion. Though there are 
many hundreds of styles 
which might be presented, 
the foregoing are ample 
for our purpose. 

It will now be in order 
to take up the construc¬ 
tion of roofs, and describe the methods by which such 
construction is obtained. 

The method of obtaining the lengths and bevels of 
















PRACTICAL EXAMPLES 


93 


rafters for ordinary roofs, such as that shown in Fig 
48, has already been given in the chapter on the steel 
square. Something has also been said regarding hip 
and valley roofs; but not enough, I think, to satisfy 
the full requirements of the workman, so I will 
endeavor to give a clearer idea of the construction of 
these roofs by employing the graphic system, instead 
of depending altogether on the steel square, though I 



earnestly advise the workman to “stick to the square.*’ 
It never makes a mistake, though the owner may in its 
application. 

A “hip roof,” pure and simple, has no gables, and 
is often called a “c :ttage :*ocf,” because of its being 
best adapted for cottages having only one, or one and 
a half, stories. The chief difficulty in its construction 
is getting the lengths and bevels of the hip or angle 
rafter and the jack or cripple rafter. To the expert 
workman, this is an easy matter, as he can readily 
obtain both lengths and bevels by aid of the square, or 
by lines such as I am about to produce. 


94 


MODERN CARPENTRY 


The illustration shown at Fig. 49 shows the simplest 
form of a hip roof. Here the four hips or diagonal 
rafters meet in the center of the plan. Another style 
of hip roof, having a gable and a ridge in the center 

of the building, is shown at 
Fig. 50. This is quite a 
common style of roof, and 
under almost every condi¬ 
tion it looks well and has 
a good effect. The plan 
shows lines of hips, valleys and ridges. 

The simplest form of roof is that known as the 
“lean-to” roof. This is formed by causing one side 
wall to be raised higher than the opposite side wall, so 
that when rafters or 
joists are laid from the 
high to the low wall a 
sloping roof is the re¬ 
sult. This style of a 
roof is sometimes called 
a “shed roof” or a 
“pent roof. ’ ’ The shape 
is shown at Fig. 51, the 
upper sketch showing 
an end view and the 
lower one a plan of the 
roof. The method of 
framing this roof, or 
adjusting the timbers 
for it, is quite obvious and needs no explanation. 
This style of roof is in general use where an annex or 
shed is built up against a superior building, hence ita 
name of “lean-to,” as it usually “leans” against the 
main building, the wall of which is utilized for 




































PRACTICAL EXAMPLES 


9 S 

high part of the shed or annex, thus saving the cost of 
the most important wall of the structure. 

Next to the “lean-to” or “shed roof” in simplicity 
comes the “saddle” or “double roof.” This roof is 
shown at Fig. 52 by the end view on the top of the fig¬ 
ure, and the plan at the bottom. It will be seen that 
this roof has a double slope, the planes forming the 
slopes are equally inclined to the horizon; the meet¬ 
ing of their highest sides makes an arris which is 
called the ridge of the roof; 
and the triangular spaces at 
the end of the walls are 
called gables. 

It is but a few years ago 
when the mansard roof was 
very popular, and many of 
them can be found in the 
older parts of the country, 
having been erected be¬ 
tween the early fifties and 
the eighties, but, for many 
¥ig,' 51 . reasons, they are now less 
used. Fig. 53 shows a roof of this kind. It is pene¬ 
trated generally by dormers, as shown in the sketch, 
and the top is covered either by a “deck roof” or a 
very flat hip roof, as shown. Sometimes the sloping 
sides of these roofs are curved, which give them a 
graceful appearance, but adds materially to their cost. 

Another style of roof is shown at Fig. This is a 
gambrel roof, and was very much in evidence in pre¬ 
revolutionary times, particularly among our Knicker¬ 
bocker ancestors. In conjunction with appropriate 
dormers, this style of roof figures prominently in what 
is known as early “coldnial style*’* It has some 




Fig. 52 , . 












96 


MODERN CARPENTRY 


advantages over the mansard. Besides these there are 
many other kinds of roofs, but it is not my purpose to 
enter largely into the matter of styles of roofs, but 
simply to arm the workman with such rules and prac¬ 
tical equipment that he 
will be able to tackle 
with success almost any 
kind of a roof that he 
may be called upon to 
construct. 

When dealing with 
the steel square I ex¬ 
plained how the lengths 
and bevels for common rafters could be obtained by 
the use of the steel square alone; also hips, purlins, 
valleys and jack rafters might be obtained by the use 
of the square, but, in order to fully equip the workman, 
I deem it necessary to present for his benefit a graphic 
method of obtaining the lengths, cuts and backing of 
rafters and purlins 
required for a hip 
roof. 

At Fig. 55, I 
show the plans of 
a simple hip roof 
having a ridge. 

The hips on the 
plan form an angle of 45 0 , or a miter, as it were. The 
plan being rectangular leaves the ridge the length of 
the difference between the length and the width of the 
building. Make cd on the ridge-line as shown, half 
the width of ab , and the angle bda will be a right angle. 
Then if we extend bd to e, making de the rise of the 
roof, ae will be the length of the hip rafter, and the 












PRACTICAL EXAMPLES 


91 

angle at x will be the plumb cut at point of hip and 
the angle at a will be the cut at the foot of the rafter. 
The angle at v shows the backing of the hip. This 
bevel is obtained as follows: Make ag and ah equal 
distances any distance will serve—then draw a line 
hg across the angle of the building, then with a center 
on ad at p, touching the line ae at s, describe a circle 
as shown by the dotted line, then draw the lines kh and 



kg y and that angle, as shown by the bevel v, will be 
the backing or bevel for the top of the hip, beveling 
each way from a center line of the hip. This rule for 
backing a hip holds good in all kinds of hips, also for 
guttering a valle}/- rafter, if the bevel is reversed. A 
hip roof where aii the hips abut each other in the cen¬ 
ter is shown in Fig. 56. T is style of roof is generally 
called a “pyramida* :ooi ’ because it has the appear¬ 
ance of a low flattened pyramid. The same rules 
governing Fig. 55 apply to this example. The bevels 
C and B show the backing 'of the hip, B showing the 































9 8 


MODERN CARPENTRY 


top from the center line 
ae\ and C showing the 
bevel as placed against 
the side of the hip, which 
is always the better way 
to work the hip. A por¬ 
tion of the hip backed is 
shown at C. The rise of 
the roof is shown at O. 

At Fig. 57 a plan of a 
roof is shown where the 
seats of the hips are not 
on an angle of 45 0 and 
where the ends and sides 
of the roof are of different 
pitches Take the'base line of the hip, ae or eg, and 
make ef perpendicular to ae, from e, and equal to 
the rise at f; make fa or fg for the length of the hip, 
by drawing the line Im at right angles to ae. This 
gives the length of the hip rafter. The backing of the 
hip is obtained in a like manner to former examples, 
only, in cases of 
this kind, there 
are two bevels for 
the backing, one 
side of the hip 
being more acute 
than the other as 
shown at D and 
E. If the hips 
are to be mitered, 
as is sometimes 
the case in roofs 
of this kind, then 


























PRACTICAL EXAMPLES 


99 



the back of the hip 
will assume the 
shape as shown by 
the two bevels at F. 
A hip roof having 
an irregular plan is 
shown at Fig. 58. 
This requires no ex¬ 
planation, as the hips and bevels are obtained in the 
same manner as in previous examples. The backing 
of the hips is shown at FG. 

An octagon roof is shown at Fig. 59, with all the 
lines necessary for getting the lcngths, bevels, and back¬ 
ing for the hips. 

The 1 i n e ax 
shows the seat 
of the hip, xe 
the rise of roof, 
and ae the 
length of hip 
and plumb cut, 
and the bevel at 
E shows the 
backing of the 
hips. 

These exam¬ 
ples will be 
quite sufficient 
to enable the 
workman to 
understand the 
general theory 
of laying out 
hip roofs. I 


g—11- 


% 


_V 



\ Fig. 59 

/ * \ i 

































loo 


modern carpentry 


may also state that to save a repetition of drawing and 
explaining the rules that govern the construction of 
hip roofs, such as I have presented serve equally well 
for skylights or similar work. Indeed, the clever 
workman will find hundreds of instances in his work 
where the rules given will prove useful. 



There are a number of methods for getting the 
lengths and bevels for purlins. I give one here which 
I think is equal to any other, and perhaps as simple. 
Suppose Fig. 60 shows one end of a hip roof, also the 
rise and length of common rafters. Let the purlin be in 
any place on the rafter, as I, and in its most com¬ 
mon position, that is, standing square with the rafter; 
then with the point b as a center with any radius, 
describe a circle. Draw two lines, ql and p?i , to touch 


















PRACTICAL EXAMPLES 


iot 


the circle p and q parallel to fb and at the points s and 
r, where the two sides of the purlin intersect, draw two 
parallel lines to the former, to cut the diagonal in m 
and k\ then G is the down bevel and F the side bevel 
of the purlin; these two bevels, when applied to the 
end of the purlin, and when cut by them, will exactly 
fit the side of the hip rafters. 

To find the cuts of a purlin where two sides are 
parallel to horizon: The square at B and the bevel at 
C will show how to draw the end of the purlin in this 
easy case. The following is universal in all posi¬ 
tions of the purlin: Let ab be the width of a square 
roof, make bf or ae one-half of the width, and make cd 
perpendicular in the middle of ef the height of the 
roof or rise* which in this case is one-third; then draw 
de and df, which are each the length of the common 
rafter. 

To find the bevel of a jack rafter against the hip, 
proceed as follows: Turn the stock of the side bevel 
at F from a around to the line iz, which will give the 
side bevel of the jack rafter The bevel at A, which is 
the top of the common rafter, is the down bevel of the 
jack rafter. 

At D the method of getting the backing of a hip 
rafter is shown the same as explained in other figures. 

There are other methods of obtaining bevels for 
purlins, but the one offered here will suffice for all 
practical purposes. 

I gave a method of finding the back cuts for jack 
rafters by the steel square, in a previous chapter. I 
give another rule herewith for the steel square: Take 
the length of the common rafter on the blade and the 
run of the same rafter on the tongue, and the blade of 
the square will give the bevel for the cut on the back 


102 


MODERN CARPENTRY 


of the jack rafter. For example, suppose the rise to 
be 6 feet and the run 8 feet, the length of the common 
rafter will be io feet. Then take io feet on the blade 
of the square, and 8 feet on the tongue, and the blade 
will give the back bevel for the cut of the jack 
rafters. 

To obtain the length of jack rafters is a very simple 
process, and may be obtained easily by a diagram, as 
shown in Fig. 6i, which is a very common method: 


First lay off half the width 
of the building to scale, as 
from A to B, the length of 
the common rafter B to C, 
and the length of the hip 
rafter from A to C. Space 
off the widths from jack 
rafter to jack rafter as shown 
by the lines I, 2, 3, and 
measure them accurately. 
Then the lines 1, 2, and 3 
will be the exact lengths of 





the jack rafters in those divisions Any number of 
jack rafters may be laid off this way, and the result 
will be the length of each rafter, no matter what may 
be the pitch of the roof or the distance the rafters are 
apart. 

A table for determining the length of jack rafters is 
given below, which shows the lengths required for 
different spacing in three pitches: 

One-quarter pitch roof: 

They cut 13 5 inches shorter each time when spaced 
12 inches. 

They cut 18 inches shorter each time when spaced 
16 inches. 








PRACTICAL EXAMPLES 103 

They cut 27 inches shorter each time when spaced 24 
inches. 

One-third pitch roof: 

They cut 14*4 inches shorter each time when spaced 
12 inches. 

They cut 19.2 inches shorter each time when spaced 
16 inches. 

They cut 28.8 inches shorter each time when spaced 
24 inches. 

One-half pitch roof: 

They cut 17 inches shorter each time when spaced 
12 inches. 



They cut 22.6 inches shorter each time when spaced 
16 inches. 

They cut 34 inches shorter each time when spaced 
24 inches. 

It is not my intention to enter deeply into a discus¬ 
sion of the proper methods of constructing roofs of all 
shapes, though a few hints and diagrams of octagonal, 
domical and other roofs and spires will doubtless be 
of service to the general workman. One of the most 
useful methods of trussing a roof is that known as a 
lattice “built-up” truss roof, similar to that shown at 
Fig. 62. The rafters, tie beams and the two main 
braces A, A, must be of one thickness—say, 2 x 4 or 
2x6 inches, according to the length of the span- 
while the minor braces are made of i-inch stuff and 



104 


MODERN CARPENTRY 


about io or 12 inches wide. These minor braces are 
well nailed to the tie beams, main braces and rafters. 
The main braces must be halved over each other at 
their juncture, and bolted. Sometimes the main 
braces are left only half the thickness of the rafters, 
then no halving will be necessary, but this method has 
the disadvantage of having the minor braces nailed to 
one side only. To obviate this, blocks maybe nailed to 
the inside of the main braces to make up the thickness 



required, as shown, and the minor braces can be nailed 
or bolted to the main brace. 

The rafters and tie beams are held together at the 
foot of the rafter by an iron bolt, the rafter having a 
crow-foot joint at the bottom, which is let into the tie 
beam. The main braces also are framed into the 
rafter with a square toe-joint and held in place with 
an iron bolt, and the foot of the brace is crow-footed 
into the tie beam over the wall. 

This truss is easily made, may be put together on 
the ground, and, as it is light, maybe hoisted in place 
with blocks and tackle, with but little trouble. This 
truss can be made sufficiently strong to span a roof 
from 40 to 75 feet. Where the span inclines to the 





PRACTICAL EXAMPLES 


greater length, the 
tie beams and raft¬ 
ers may be made of 
built-up timbers, but 
in such a case the 
tie beams should 
not be less than 
6 x io inches, nor 
the rafters less than 
6x6 inches. 

Another style of 
roof altogether is 
shown at Fig. 63. 
This is a self-sup¬ 
porting roof, but is 
somewhat expensive 
if intended for a 
building having a 
span of 30 feet or 
less. It is fairly 
well adapted for 
halls or for country 
churches, where a 
high ceiling is re¬ 
quired and the span 
anywhere from 30 
to 50 feet over all. 
It would not be safe 
to risk a roof of this 
kind on a building 
having a span more 
than 50 feet. The 
main features of this 
roof are: (1) having 


105 










































io6 


MODERN CARPENTRY 


collar beams, (2) truss bolts, and (3) iron straps at the 
joints and triple bolts at the feet. 

I show a dome and the manner of its construction at 
Fig. 64. This is a fine example of French timber 
framing. The main carlins are shown at a , b , c, d 
and Nos. 1 and 2, and the horizontal ribs are also 
shown in the same numbers, with the curve of the 
outer edge described on them. These ribs are cut in 
between the carlins or rafters and beveled off to suit. 
This dome may be boarded over either horizontally or 
with boards made into “gores” and 
laid on in line with the rafters or 
carlins. 

The manner of framing is well 
illustrated in Nos. 3 and 4 in two 
ways, No. 3 being intended to form 
the two principal trusses which 
stretch over the whole diameter, 
while No. 4 may be built in between 
the main trusses. 

The illustrations are simple and 
clear, and quite sufficient without 
further explanation. 

Fig. 65 exhibits a portion of the dome of St. Paul’s 
Cathedral, London, which was designed by Sir Chris¬ 
topher Wren The system of the framing of the 
external dome of this roof is given. The internal 
cupola, AAi, is of brick-work, two bricks in thickness, 
with a course of bricks 18 inches in length at every five 
feet of rise. These serve as a firm bond. This dome 
was turned upon a wooden center, whose only support 
was the projections at the springing of the dome, 
which is said to have been unique. Outside the brick 
cupola, which is only alluded to in order that the 












PRACTICAL EXAMPLES 


107 


description may be the more intelligible, rises a brick¬ 
work cone B. A portion of this can be seen, by a 
spectator on the floor of the cathedral, through the 
central opening at A. The timbers which carry the 
external dome rest upon this conical brickwork. The 
horizontal hammer beams, C, D, E, F, are curiously 
tied to the corbels, G, H, I, K, by iron cramps, well 
bedded with lead into the 



of the dome is made, pass among the roof trusses. 
The dome has a planking from the base upwards, and 
hence the principals are secured horizontally at a little 
distance from each other. The contour of this roof is 
that of a pointed dome or arch, the principals being 
segments of circles; but the central opening for the 
lantern, of course, hinders these arches from meeting 
at a point. The scantling of the curved principals is 
10x II y 2 inches at the ba$e, decreasing to 6x6 inches 









io8 


MODERN CARPENTRY 


at the top. A lantern of Portland stone crowns the 
summit of the dome. The method of framing will be 
clearly seen in the diagram. It is in every respect an 
excellent specimen of roof construction, and is worthy 
of the genius and mathematical skill of a great work¬ 
man. 

With the rules offered herewith for the construction 
of an octagonal spire, I close the subject 
of roofs: To obtain bevels and lengths of 
braces for an octagonal spire, or for a 
spire of any number of sides, let AB, 
Fig. 66, be one of the sides. Let AC and 
BC be the seat line of hip. Let AN be 
the seat of brace. Now, to find the posi¬ 
tion of the tie beam on the hips so as to 
be square with the boarding, draw a line 
through C, square with AB, indefinitely. 
From C, and square with EC, draw CM, 
making it equal to the height. Join EM. 
Let OF be the height of the tie beam. 
At F draw square with EM a line, which 
produce until it cuts EC prolonged at G. 
Draw CL square with BC. Make CL in 
length equal to EM. Join BL, and make NH equal to 
OF. From G draw the line GS parallel with AB, cut¬ 
ting BC prolonged, at the point S; then the angle at H 
is the bevel on the hip for the tie beam. For a bevel 
to miter the tie beam, make FV equal ON. Join VX; 
then the bevel at V is the bevel on the face. For the 
down bevel see V, in Fig. 67. To find the length of 
brace, make AB, Fig. 67, equal to AB, Fig. 66. Make 
AL and BL equal to BL, Fig. 66. Make BP equal to 
BH. Join AP and BC, which will be the length of the 
brace. The bevels numbered 1, 3, 5 and 7 are all to be 





PRACTICAL EXAMPLES 


209 

used, as shown on the edge of the brace. No. 1 is to 
be used at the top above No. 5. For the bevel on the 
face to miter on the hip, draw AG, Fig. 66, cutting BS 
at J. Join JH. Next, in Fig. 68, make AP equal AP, 
Fig. 67, and make AJ equal to AJ, Fig. 66. Make 
PJ equal to JH, Fig. 66, and make PI equal to HI. 
Join AI; then the bevel marked No. 5 will be correct 
for the beam next to the hip, and the bevel marked 
No. 6 will be correct for the top. Bevel No. 2 in this 
figure will be correct tor the beam next to the plate. 
The edge of the brace is to correspond with the 
boarding. 

A few examples of scarfing tim¬ 
ber are presented at Figs. 69, 70, 71 
and 72. The example shown at 
Fig. 69 exhibits a method by 
which the two ends of the timber 
are joined together with a step- 
splice and spur or tenon on end, it 
being drawn tight together by the 
keys, as shown in the shaded part. Fig. 70 is a similar 
joint though simpler, and therefore a better one; A, A 
are generally joggles of hardwood, and not wedged 
keys, but the latter are preferable, as they allow of 
tightening up. The shearing used along BF should be 
pine, and be not less than six and a half times BC; 
and BC should be equal to at least twice the depth of 
the key. The shear in the keys being at right angles 
to the grain of the wood, a greater stress per square 
inch of shearing area can be put upon them than 
along BF, but their shearing area should be equal in 
strength to the other parts of the joint; oak is the 
best wood for them, as its shearing is from four to five 
times that of pine. 




130 


MODERN CARPENTRY 


Scarfed joints with bolts and indents, such as that 
shown at Fig. ?i 9 are about the strongest of the kind. 
From this it will be seen that the strongest and most 
economical method in every way, in lengthening ties, 
is by adoption of the common scarf joint, as shown at 
Fig. 71, and finishing the scarf as there represented. 

The carpenter meets with many conditions when 
timbers of various kinds have to be lengthened out 




and spliced, as in the case of wall plates, etc., where 
there is not much tensile stress. In such cases the 
timbers may simply be halved together and secured 
with nails, spikes, bolts, screws or pins, or they may 
































PRACTICAL EXAMPLES 


in 


he halved or beveled as shown in Fig. 72, which, when 
ioarded above, as in the case of wall plates built in 
the wall, or as stringers on which partitions are set, or 
joint beams on which the lower edges of the joists rest, 
will hold good together. 

Treadgold gives the following rules, based upon the 
relative resistance to tension, crushing and shearing 
of different woods, for the proportion which the length 
or overlap of a scarf should bear to the depth of the 
tie: 


Without 

bolts 

Oak, ash, elm, etc. . . 6 

Pine and similar woods . 12 


With 

bolts 


3 

6 


With bolts 
and indents 
2 

4 


There are many other kinds of scarfs that will occur 
to the workman, but it is thought the foregoing may 
be found useful on special 
occasions. 

A few examples of odd 
joints in timber work will 
not be out of place. It 
sometimes happens that 
cross-beams are required 
to be fitted in between 
girders in position, as in 
renewing a defective one, 
done, and a mortise and tenon joint is used, a chase 
has to be cut leading into the mortise, as shown in the 
horizontal section, Fig 73. By inserting the tenon at 
the other end of the beams into a mortise cut so as to 
allow of fitting it in at an angle, the tenon can be slid 
along the chase b into its proper position. It is better 
in this case to dispense with the long tenon, and, if 
necessary, to substitute a bolt, as shown in the sketch. 
A mortise of this kind is called a chase mortise , but an 



and when this has to be 







II* 


MODERN CARPENTRY 




Ffc.74 


> 


iron shoe made fast to the girder forms a better means 
of carrying the end of a cross-beam. The beams can 
be secured to the shoe with bolts or other fastenings. 

To support the end of a horizontal beam or girt on 
the side of a post, the joint shewn in Fig. 74 may be 
used where the mortise for 
the long tenon is placed, to 
weaken the post as little as 
possible, and the tenon made 
about one-third the thickness 
of the beam on which it is cut. 
The amount of bearing the 
beam has on the post must 
greatly depend on the work it 
has to do. A hardwood pin 
can be passed through the 
cheeks of the mortise and the tenon as shown to keep 
the latter in position, the holes being draw-bored in 
erder to bring the shoulders of the tenon tight home 
against the post, but care must be taken not to overdo 
the draw-boring or the wood at the end of the tenon 
will be forced out by the 
pin. The usual rule for 
draw-boring is to allow a 
quarter of an inch draw in 
soft woods and one-eighth 
of an inch for hard woods. 





These allowances may seem rather large, but it must 
be remembered that both holes in tenon and mortise 
will give a little, so also will the draw pin itself unless 
it is of iron, an uncommon circumstance. 

Instead of a mortise and tenon, an iron strap or & 
screw bolt or nut may be used, similiar to that shown 
in Fig. 75. 













PRACTICAL EXAMPLES 


IT 3 

The end of the beam may also be supported on a 
block which should be of hardwood, spiked or bolted 

on to the side of the 
post, as at A and B, 
Fig. 76. The end of 
the beam may either 
be tenoned into the 
post as shown, or it 
may have a shoulder, 
with the end of the 
beam beveled, as 
shown at A. 

Heavy roof tim¬ 
bers are rapidly giv¬ 
ing place to steel, but 
there yet remain 
many cases w h e r e 
timbers will remain employed and the old method of 
framing continued. The use of iron straps and bolts 
in fastening timbers together or for trussing purposes 
will never perhaps become obsolete, therefore a knowl¬ 
edge of the proper use of 
these will always remain 
valuable. 

Heel straps are used to 
secure the joints between 
inclined struts and hori¬ 
zontal beams, such as the 
joints between rafters and 
beams. They may be placed either so as merely to 
hold the beams close together at the joints, as in Fig. 
77, or so as to directly resist the thrust of the inclined 
strut and prevent it from shearing off the portion of 
the horizontal beam against which it presses. Straps 














U4 


MODERN CARPENTRY 


of the former kind are sometimes called kicking-straps . 
The example shown at Fig. 77 is a good form of strap 
for holding a principal rafter down at the foot of the 
tie beam. The screws and nuts are prevented from 
sinking into the wood by the bearing plate B, which 
acts as a washer on which the nuts ride when tighten¬ 
ing is done. A check plate is also provided under¬ 
neath to prevent 
the strap cutting 
into the tie beam. 

At Fig. 78 I show 
a form of joint 
often used, but it 
represents a diffi¬ 
culty in getting 
the two parallel 
abutments to take 
their fair share of 
the work, both 
from want of accu¬ 
racy in workman¬ 
ship as well as 
from the disturb¬ 
ing influence of 
shrinkage. In 
making a joint of this sort, care must be taken that 
sufficient wood is left between the abutments and the 
end of the tie beam to prevent shearing. A little 
judgment in using straps will often save both time 
and money and yet be sufficient for all purposes, 

I show a few examples of strengthening and trussing 
joints, girders, and timbers at Fig 79. The diagrams 
need no explanation, as they are self-evident 

It would expand this book far beyond the dimensions 











PRACTICAL EXAMPLES n S 

awarded me, to even touch on all matters pertaining 
to carpentry, including bridges, trestles, trussed gird¬ 
ers and trusses generally, so I must content myself 








Fig. 79i 


with what has already been given on the subject of 
carpentry, although, as the reader is aware, the subject 
is only surfaced. 
































r * 


*• 














% 











t 





« 
















* 








PART III 


JOINER’S WORK 
CHAPTER I 

KERFINC, RAKING MOULDINGS, HOPPERS AND SPLAYS 

This department could be extended indefinitely, as 
the problems in joinery are much more numerous than 
in carpentry, but as the limits of this book will not 
permit me to cover the whole range of the art, even if 

I were competent, I 
must be contented 
with dealing with 
those problems the 
workman will most 
likely be confronted 
with in his daily oc¬ 
cupation. 

First of all, I give several methods of “kerfing,” for 
few things puzzle the novice more than this little 
problem. Let us suppose any circle around which it 
is desired to bend a piece of stuff to be 2 inches larger 
on the outside than on the inside, or in other words, 
the veneer is to be I inch thick, then take out as many 
saw kerfs as will measure 2 inches. Thus, if a saw 
cuts a kerf one thirty-second of an inch in width, then 
it will take 64 kerfs in the half circle to allow for the 


0 







MODERN CARPENTRY 


118 


veneer to bend around neatly. The piece being 
placed in position and bent, the kerfs will exactly 
close. 

Another way is to saw one kerf near the center of 
the piece to be bent, then place it on 
the plan of the frame, as indicated in the 
sketch and bend it until the kerf closes. 
The distance, DC, Fig. I, on the line DB, 
will be the space between the kerfs neces¬ 
sary to complete the bending. 

In kerfing the workman should be care¬ 
ful to use the same saw throughout, and to 
!/1 cut exactly the same depth every time, and 
the spaces must be of equal distance. In 
diagram Fig. I, DA shows the piece to 
be bent, and at O the thickness of the 
stuff is shown, also path of the inside and 
outside of the circle. 

Another, and a safe method of kerfing 
is shown at Fig. 
2, in which it is 
desired to bend 
a piece as 
shown, and 
which is in¬ 
tended to be 
secured at the 
ends. Up to A 
is the piece to 
be treated. 
First gauge a line on about one-eighth inch back from 
the face edges, and try how far it will yield when the 
first cut is made up to the gauge line, being cut perfectly 
straight through from side to side, then place the work 













JOINER’S WORK 


119 

cn a flat board and try it gently until the kerf closes, 
and it goes as far as is shown at A, which is the first 
cut, B representing the second. Those are the dis¬ 
tances the kerfs require to be placed apart to complete 
the curve. Try the work as it progresses. This eases 
the back of ;t and makes it much easier done when the 
whole cuts are finished. Now make certain that the 
job will fold to the curve, then fill them all with hot 
glue and proceed to fix. The plan shown here is a 
half semi, and 
may be in excess 
of what is wanted, 
but the principle 
holds good. 

Another method 
is shown at Fig. 3 
for determining 
the number and 
distances apart of the saw kerfs required to bend a 
board round a corner. The board is first drawn in 
position and a half of it divided into any number of 
equal parts by radii, as I, 2, 3, 4, 5, 6. A straight 
piece is then marked off to correspond with the divi¬ 
sions on the circular one. By this it is seen that the 
part XX must be cut away by saw kerfs in order to let 
the board turn round. It therefore depends upon the 
thickness of the saw for the number of kerfs, and when 
that is known the distances apart can be determined as 
shown on the right in the figure. Here eight kerfs are 
assumed to be requisite. 

To make a kerf for bending round an ellipse, such as 
that shown at Fig. 4, proceed as shown, CC and 00 
being the distances for the kerfs; 2 to 2 and 2 to 3 are the 
lengths of the points EF, while BB is the length of the 









120 


MODERN CARPENTRY 


points EE, making 1 the. whole head piece in one. In 
case it is necessary to joint D, leave the ends about 8 
inches longer than is necessary, as shown by N in the 



sketch, so that should a breakage occur this extra 
length may be utilized. 

It is sometimes necessary to bend thick stuff around 
work that is on a rake, and when this is required, all 
that is necessary is to run in the kerfs the angle of the 
rake whatever that may be, as 
shown at Fig. 5. This rule holds 
good for all pitches or rakes. 

Fig. 6 shows a very common 
way of obtaining the distance 
to place the kerfs. The piece 
to be kerfed is shown at C; 
now make one at E; hold firm 
the lower part of C and bend 




















JOINER’S WORK 


I9X 


the upper end on the circle F until the kerf is closed. 
The line started at E and cutting the circumference of 
the circle indicates at the circumference the distance 
the saw kerfs will be apart. Set the dividers to this 
space, and be¬ 
ginning at the 
center cut, 
space the piece 
to be kerfed 
both ways. 

Use the same 
saw in all cuts 
and let it be 
clean and keen, 
with all dust 
well cleaned 
out. 

To miter 
mouldings, 

where straight lines must merge into lines having a 
curvature as in Figs. 7 and 8: In all cases, where a 
straight moulding is intersected with a curved mould¬ 
ing of the same profile at whatever angle, the miter is 
necessarily other than a straight line. The miter line 

is found by the intersec¬ 
tion of lines from the 
several points of the pro¬ 
file as they occur respect¬ 
ively in the straight and 
the curved mouldings. 
In order to find the miter 
between two such mould- 
irgs, first project lines 
from all of the points of 


















122 


MODERN CARPENTRY 


the profile indefinitely to the right, as shown in the 
elevation of the sketch. Now, upon the center line of 
the curved portion, or upon any line radiating from 
the center around which the curved moulding is to be 

carried, set off the 
several points of 
the profile, spac¬ 
ing them exactly 
the same as they 
are in the eleva¬ 
tion of the straight 
moulding. Place 
one leg of the 
dividers at the 
center of the cir¬ 
cle, bringing the other leg to each of the several points 
upon the curved moulding, and carry lines around the 
curve, intersecting each with a horizontal line from 
the corresponding point of the level moulding. The 
dotted line drawn through the intersections at the 
miter shows what 
must be the real 
miter line. 

Another odd miter¬ 
ing of this class is 
shown in Fig. 9. In 
this it will be seen 
that the plain faces 
of the stiles and 
circular rail form 
junctions, the mould¬ 
ings all being mi¬ 
tered. The miters 
are curved in order 

























JOINER’S WORK 


”3 

to have all the members of the mouldings merge in 
one another without overwood. Another example is 
shown at Fig. io, where the circle and mouldings 
make a series of panels. These examples are quite 
sufficient to enable the 
workman to deal effect¬ 
ively with every prob¬ 
lem of this kind. 

The workman some¬ 
times finds it a little 
difficult to lay out a hip 
rafter for a veranda that 
has a curved roof. A 
very easy method of finding the curve of the hip is 
shown at Fig. u. Let AB be the length of the angle 
or seat of hip, and CO the curve; raise perpendicular 

on AB, as shown, 
same as those on 
DO, and trace 
through the points 
obtained, and the 
thing is done. 

Another simple 
way of finding the 
hip for a single curve 
is shown at Fig. 12; 
AB represents the 
curve given the com¬ 
mon rafter. 

Now lay off any number of lines parallel with the 
seat from the rise, to and beyond the curve AB, as 
shown, and for each inch in length of these lines 
(between rise and curve), add T % of an inch to the 
same line to the left of the curve, and check. After 

























MODERN CARPENTRY 


124 


all lines have thus been measured, run an off-hand 
curve through the checks, and the curve will represent 
the corresponding hip at the center of its back. 

To find the bevel 
or backing of the hip 
to coincide with the 
plane of the common 
rafter, measure back 
on the parallel lines 
to the right of the 
curve one-half the 
thickness of the hip 
and draw another 
curve, which will be 
the lines on the side 
to trim to from the 
center of the back. 
A like amount must 
be added to the 
plumb cut to fit the 
corner of deck. Pro¬ 
ceed in like manner 
for the octagon hip, 
but instead of adding 
t 5 5 , add of an inch 
as before described. 

[Wh i 1 e this is 
worked out on a giv¬ 
en rise and run for the 
rafter, the rule is applicable to any rise or run, as the 
workman will readily understand.] 

A more elaborate system for obtaining the curve of a 
hip rafter, where the common rafters have an ogee or 
concave and convex shape, is shown at Fig. 12^. This 
















JOINER’S WORK 


125 


is a very old method, and is shown—with slight varia¬ 
tions—in nearly all the old works on carpentry and 
joinery. Draw the seat of the common rafter, AB, 
and rise, AC. Then draw the curve of the common 
rafter, CB. Now divide the base line, AB, into any 
number of equal spaces, as 1, 2, 3, 4, 5, etc., and draw 
perpendicular lines to construct the curve CB, as 1 o, 
2 0, 30, 40, etc. Now draw the seat of the valley, or 
hip rafter, as BD, and continue the 
perpendicular lines referred to until 
they meet BD, thus establishing the 
points 10, II, 12, 13, 14, etc. From 
these points draw lines at right 
angles to BD, making 10 x equal in 
length to 1 o, and 11 x equal to 2 oj 

Figs. 13 , 



also 12 x equal to 3 o, and so on. When this has been 
done draw through the points indicated by x the 
curve, which is the profile of the valley rafters. 

Another method, based on the same principles as 
Fig. 12%, is shown at Fig. 13. Let ABCFED represent 
the plan of the roof. FCG represents the profile of the 
wide side of common rafter. First divide this common 
rafter, GC, into any number of parts—in this case 6. 




















126 


MODERN CARPENTRY 


Transfer these points to the miter line EB, or, what is. 
the same, the line in the plan representing the hip 
rafter From the points thus established at E, erect 
perpendiculars indefinitely With the dividers take 
the distance from the points in the line FE, measur¬ 
ing to the points in the profile GC, and set the same 
off on corresponding lines, measuring from EB, thus 
establishing the points I, 2, etc.; then a line traced 

through these 
points will be th^ 
required hip rafter. 

For the com¬ 
mon rafter, on the 
narrow side, con¬ 
tinue the lines from 
EB parallel with 
the lines of the 
plan DE and AB. 
Draw AD at right 
angles to these 
lines. With the 
dividers, as before, measuring from FE to the points 
in GC, set off corresponding distances from AD, thus 
establishing the points shown between A and H. A 
line traced through the points thus obtained will be 
the line of the rafter on the narrow side. 

These examples are quite sufficient to enable the 
workman to draw the exact form of any rafter no mat¬ 
ter what the curve of its face may be, or whether it is 
for a veranda hip, or an angle bracket, for a cornice 
or niche. 

Another class of angular curves the workman will 
meet with occasionally, is that when raking mould¬ 
ings are used to work in level mouldings, as fp* 















JOINER’S WORK 


127 


instance, a moulding down a gable that is to miter. 
The figures shaded in Fig. 14 represent the mould- 
ing in its various phases and angles. Draw the out¬ 
line of the common level moulding, as shown at F, in 
the same position as if in its place on the building. 
Draw lines through as many prominent points in the 
profile as may be convenient, parallel with the line of 
rake. From the same points in the moulding draw ver¬ 
tical lines, as shown by iH, 2, 3, 4 and 5, etc. From 
the point I, square with the lines of the rake, draw iM, 



as shown, and from 1 as center, with the dividers 
transfer the divisions 2, 3, 4, etc., as shown, and from 
the points thus obtained, on the upper line of the rake 
draw lines parallel to iM. Where these lines intersect 
with the lines of the rake will be points through which 
the outline C may be traced. 

In case there is a moulded head to put upon a raking 













128 


MODERN CARPENTRY 


gable, the moulding D shown at the right hand must 
be worked out for the upper side. The manner in 
which this is done is self-evident upon examination 
of the drawing, and therefore needs no special 
description. 

A good example of a raking moulding and its appli¬ 
cations to actual work is shown in Fig. 15, on a differ¬ 
ent scale. The ogee moulding at the lower end is the 
regular moulding, while the middle line, ax^ shows 
the shape of the raking moulding, and the curve on 



the top end, cdo , shows the face of a moulding that 
would be required to return horizontally at that point. 
The manner of pricking off these curves is shown by 
the letters and figures. 

At Fig. 16 a finished piece of work is shown, where 
this manner of work will be required, on the returns. 

Fig. 17 shows the same moulding applied to a 
curved window or door head. The manner of pricking 
the curve is given in Fig. 18. 

At No. 2 draw any line, AD, to the center of the 


















JOINER’S WORK 


129 


pediment, meeting the upper edge of the upper fillet 
in D, and intersecting the lines AAA, aaa , bbb, ccc y 



aa y bb } cc, BB, EE, tangents to their respective arcs; 
































i 3 o MODERN CARPENTRY 

on the tangent line DE, from D, make D d, De, D/ 
DE, respectively equal to the distances D d, De, Df, 
DE on the level line DE, at No. i. Through the 
points d, e, f E, draw da, eb, fc, EB, then the curve 
drawn through the points A, a, b, c, B, will be the sec¬ 
tion of the circular moulding. 

Sometimes mouldings for this kind of work are made 

of thin stuff, 
and are bev¬ 
eled on the 
back at the 
bottom in 
such a man¬ 
ner that the 
top portion 
of the mem¬ 
ber hangs 
over, which 
gives it the 
appearance 
of being 
solid. 
Mo u 1 dings 
of this kind 
are called 

“spring mouldings,” and much care is required in 
mitering them. This should always be done in a 
miter box, which must be made for the purpose; often 
two boxes are required, as shown in Figs. 19-22. The 
cuts across the box are regular miters, while the angles 
down the side are the same as the down cut of the 
rafter, or plumb cut of the moulding. When the box 
is ready, place the mouldings in it upside down, keep¬ 
ing the moulded side to the front, as seen in Fig. 20^ 


1 = 


=7= 


\ 

Pig- 19 , 

/ 

) 

f= 

-1—- 



1 Fig. 21, 

i 

) 


20 


Fig. 2X 





























JOINER’S WORK 


131 


making sure that the level of the moulding at c fits 
close to the side of the box. 

To miter the rake mouldings together at the top, 
the box shown in Fig. 21 is used. The angles on the 
top of the box are 
the same as the 
down bevel at the 
top of the rafter, the 
sides being sawed 
down square. Put 
the moulding in the 
box, as shown in 
Fig. 22, keeping the 
bevel at c flat on the 
bottom of the box, 
and having the 
moulded side to the 
front, and the miter 
for the top is cut, 
which completes the 
moulding for one 
side of the gable. 

The miter for the 
top of the moulding 
for the other side of 
the gable may then 
be cut. 

When the rake 
moulding is made of 
the proper form these 
boxes are very con¬ 
venient; but a great 
deal of the machine- 
made mouldings are 




















MODERN CARPENTRY 



not of the proper form to fit. In such cases the 
moulding should be made to suit, or they come bad; 
although many use the mouldings as they come from 
the factory, and trim the miters so as to make them 

do. 

The instructions given, however, in Figs. 13, 14, 15 

and 18 will enable 
the workman to 
make patterns for 
what he requires. 

While the 
“angle bar” is not 
much in vogue at 
the present rime, 
the methods by 
which it is ob¬ 
tained, may be ap¬ 
plied to many pur¬ 
poses, so it is but 
proper the method 
should be em¬ 
bodied in this 
work. In Fig. 23, 
B is a common 
sash bar, and C is 
the angle bar of 
the same thick¬ 
ness. Take the raking projection, 11, in C, and set the 
foot of your compass in 1 at B, and cross the middle 
of the bar at the other 1; then draw the points 2, 2, 3, 3, 
etc., parallel to 11, then prick your bar at C from the 
ordinates so drawn at B, which, when traced, will give 
the angle bar. 

This is a simple operation, and may be applied to 














JOINER’S WORK 


133 


many other cases, and for enlarging or diminishing 
mouldings or other work. 

The next figure, 24, gives the lines for a raking 
moulding, such as a cornice in a room with a sloping 
coiling As may be 
oeen from the dagram 
the three sections 
;hown are drawn equal 
n thickness to miter at 
.he angles of the room. 

The construction 
should be easily under¬ 
stood When a straight 
moulding is mitered 
with a curved one the 
line of miter is some¬ 
times straight and sometimes curved, as seen at Fig. 
18, and when the mouldings are all curved the miters 
are also straight and curved, as shown in previous 
examples. 

If it is desired to make a cluster column of wood, it 
is first necessary to make a standard or core, which must 
have as many sides as there are to be faces of columns. 

Fig. 25 shows how the work is 
done. This shows a cluster of 
four columns, which are nailed to 
a square standard or core. Fig. 
26 shows the base of a clustered 
column. These are blocks turned 
in the lathe, requiring four of 
them for each base, which are cut 
and mitered as shown in Fig. 25. 
The cap, or capital, is, of course, 
cut in' the same manner. 








134 


MODERN CARPENTRY 


Laying out lines for hopper cuts is often puzzling, 
and on this account I will devote more space to this 
subject than to those requiring less explanations. 

Fig. 27 shows an isometric view of three sides of a 
hopper. The fourth side, or end, is ['purposely left 
out, in order to show the exact build of the hopper. 
It will be noticed that AC and EO show the end of the 

work as squared 
up from the bot¬ 
tom, and that BC 
shows the gain of 
the splay or flare. 
This gives the idea 
of what a hopper 
is, though the 
width of side and 
amount of flare 
may be any meas¬ 
urement that may 
be decided upon. 
The difficulty in 
this work is to get 
the proper lines for the miter and for a butt cut. 

Let us suppose the flare of the sides and ends :o ho 
as shown at Fig. 28, though any flare or inclination 
will answer equally well. This diagram and the plan 
exhibit the method to be employed, where the sides 
and ends are to be mitered together. To obtain the 
bevel to apply for the side cut, use A' as center, B' as 
radius, and CDF' parallel to BF. Project from B to 
D parallel to XY. Join AD, which gives the bevel 
required, as shown. If the top edge of the stuff is to be 
horizontal, as shown at B'G', the bevel to apply to the 
edge will be simply as shown in plan by BG; but if 






JOINER’S WORK 


*35 


the edge of the stuff is to be square to the side, as 
shown at B'C', Fig. 29, the bevel must be obtained as 
follows: Produce EB' to D', as indicated, Fig. 29. 
With B as center, describe the arc from C', which 
gives the point D. Project down from D, making DF 



parallel to CC, as shown. Project from C parallel to 
XY. This will give the point D. Join BD, and this 
will give the bevel line required. At A, Fig. 31, is 
shown the application of the bevel to the side of the 
stuff, and at B the application of the bevel to the edge 
of the stuff. When the ends butt to the sides, as indi¬ 
cated at H, Fig. 30, the bevel, it will be noticed, is 
obtained in a similar manner to that shown at Fig. 28. 
It i6 not often that simply a butt joint is used between 

























MODERN CARPENTRY 


136 


the ends and sides, but the ends are usually housed 
into the sides, as indicated by the dotted lines shown 
at H, Fig. 30. 

Another system, which was first taught by the cele¬ 
brated Peter Nicholson, and afterwards by Robert 

Riddell, o l 
Philadel¬ 
phia, is ex¬ 
plained in 
the follow- 
i n g : T h e 
i 1 1 u s t r a - 
tion shown at 
Fig. 32 is in- 
tendedto 
show how to 
find the lines 
for cutting 
butt joints 
for a hopper. 
Construct a 
right angle, 
as A, B, C, 
Fig. 32, con¬ 
tinue A, B 
pastK. From 
K, B make 
the inclination of the sides of the hopper, 2, 3. 

Draw 3, 4 at right angles with 3, 2; take 3 as center, 
and strike an arc touching the lower line, cutting in 4. 
Draw from 4, cutting the miter line in 5; from 5 square 
draw a line cutting in 6, join it and B; this gives bevel 
W, as the direction of cut on the surface of sides. To 
find the butt joint, take any two points, A, C, on the 

























JOINER’S WORK 


137 


right angle, equally distant from B, make the angle 
B, K, L, equal that of 3, K, L, shown on the left; from 
B draw through point L; now take C as a center, and 
strike an arc, touching line BL. From A draw a 
line touching the arc at H, and cutting the extended 
line through B 
in N, thus fixing 
N as a point. 

Then by draw- 
i n g from C 
through N, we 
get the bevel 
X for the butt 
joint. Joints 
on the ends of 
timbers running 
horizontally in 
tapered framed 
structures, when 
the plan is 
square and the 
inclinations 
equal, may be 
found by this 
method. 

The backing 

of a hip rafter may also be obtained by this method, as 
shown at J, where the pitch line is used as at 2, 3, 
which would be the inclination of the roof. 

The solution just rendered is intended only for hop¬ 
pers having right angles and equal pitches or splays, 
as hoppers having acute or obtuse angles, must be 
treated in a slightly different way. 

Let us suppose a butt joint for a hopper having an 





















13* 


MODERN CARPENTRY 


acute angle, such as shown at A, B, C, Fig. 33, and 
with an inclination as shown at 2, 3. Take any two 
points, A, C, equaliy distant from B. Join A, C, 
bisect this line in P, draw through P, indefinitely. 
Find a bevel for the side cut by drawing 3, 4, square 
with 2, 3; take 3 as a center, and strike an arc, touch¬ 
ing the lower line cutting in 4; draw from 4, cutting 



the miter line in 5, and from it square draw a line 
cutting in 6. Join 6, B, this gives bevel W, for direc¬ 
tion of cut on the surface of inclined sides. 

The bevel for a butt joint is found by drawing C, 8, 
square with A, B; make the angle 8, K s L, equal that 
of 3, K, L, shown on the left. Draw from 8 through point 
L; take C as a center and strike an arc touching the 
line 8, L; draw from A, touching the arc at D, cutting 











JOINER’S WORK 


i39 


the line from P, in D, making it a point, then by 
drawing from C, through D, we get the bevel X for 
the butt joint. 

As stated regarding the previous illustration, the 
backing for a hip in a roof having the pitch as shown 
at 2, 3, may be found at the bevel J. The same rule 



also applies to end joints on timbers placed in a hori¬ 
zontal double inclined frame, having an acute angle 
same as described. 

Having described the methods for finding the butt 
joints in right-angled and acute-angled hoppers, it will 
be proper now to define a method for describing an 
obtuse-angled hopper having butt joints. 

Let the inclination of the sides of the hopper be 






MODERN CARPENTRY 


140 

exhibited at the line 2, 3, and the angle of the obtuse 
corner of the hopper at A, B, C, then to find the joint, 
take any two points, A, C, equally distant from B, 
join these points, and divide the line at P. Draw 
through P and B, indefinitely. At any distance below 
the side A, B, draw the line 2, 6; make 3, 4, square 
with the inclination. From 3, as a center, describe 
an arc, touching the lower line and cutting in 4; from 
4 draw to cut the miter line in 5, and from it square 



down a line cutting in 6, join 6, B, and we get the 
bevel W, for cut on surface sides. 

The bevel for the butt joint is found by drawing C, 
D, square with B, A, and making the angle D, K, L 
equal to that of 3, K, L on the left. From C, as a 
center, strike an arc, touching the line D, L; then 
from A draw a line touching the arc H. This line 
having cut through P, in N, fixes N as a point, so that 
by drawing C through N an angle is determined, in 
which is bevel X for the butt joint. 






JOINER’S WORK 


141 

To obtain the bevels or miters is a simple matter to 
one who has mastered the foregoing, as evidenced by 
the following: 

Fig. 34 shows a right-angled hopper; its sides may 
stand on any inclination, as AB. The miter line, 



2, W, on the plan, being fixed, draw B, C square with 
the inclination. Then from B, as center, strike an arc, 
touching the base line and cutting in CD. From CD 
draw parallel with the base line, cutting the miters in 
F and E; and from these points square down the lines, 
cutting in 3 and 4. From 2 draw through 3; this gives 



bevel W for the direction of cut on the surface sides. 
Now join 2, 4, this gives bevel X to miter the edges, 
which in all cases must be square, in order that bevels 
may be properly applied. 

Fig. 35 shows a plan forming an acute-angled hop- 











142 


MODERN CARPENTRY 


per, the miter line being 2, W. The sides of this plan 
are to stand on the inclination AB. Draw BC square 
with the inclination, and from B, as center, strike an 
arc, touching the base line and cutting in CD. Draw 
from CD, cutting the miter line at E and F; from these 
points square down the lines, cutting in 3 and 4. From 
2 draw through 4, which will give bevel W to miter 
the edges of sides. Now join 2, 3, which gives bevel 
X for the direction of cut on the surface of sides. 

Fig. 36 shows an obtuse-angled hopper, its miter line 
on the plan being 2 W, and the inclination of sides 



AB. Draw BC square with the inclination, and from 
B as center strike an arc, touching the base line and 
cutting CD. Draw from CD, cutting the miter in F 
and E. From these points square down the lines, cut¬ 
ting the base; then by drawing from 2 through the 
point below E, we get bevel W for the direction of 
cuts on the surface of sides, and in like manner the 
point below F being joined with 2, gives bevel X to 
miter the edges. 

It will be noticed that the cuts for the three differ¬ 
ent angles are obtained on exactly the same principle, 
without the slightest variation, and so perfectly sim¬ 
ple as to be understood by a glance at the drawing. 
The workman will notice that in each of the angles a 






JOINER’S WORK 



. 37 . 


Sicfe 






































144 


MODERN CARPENTRY 


line from C, cutting the miter, invariably gives a direc¬ 
tion for the surface of sides, and the line from D 
directs the miter on their edges. 

Unlike many other systems employed, this one meets 
all and every condition, and is the system that has 
been employed by high class workmen and millwrights 
for ages. 

One more example on hopper work and I am done 
with the subject: Suppose it is desired to build a 
hopper similar to the one shown at Fig. 37, several 


new conditions 
will be met with, 
as will be seen by 



an examination of 


the obtuse and 
acute angles, L 
and P. In order to 
work this out 
right make a 
diagram like 


that shown at Fig. 38, where the line AD is the given 
base line on which the slanting side of hopper or box 
rises at any angle to the base line, as CB, and the 
total height of the work is represented by the line 
B, E. By this diagram it will be seen that the hori¬ 
zontal lines or bevels of the slanting sides are indi¬ 
cated by the bevel Z. 

Having got this diagram, which of course is not 
drawn to scale, well in hand, the ground plan of the 
hopper may be laid down in such a shape as desired, 
with the sides, of course, having the slant as given in 

Fig 38. 

Take T2, 3S, Fig. 37, as a part of the plan, then set 
off the width of sides equal to C, B, as shown in Fig. 38. 






JOINER’S WORK 


*45 


These are shown to intersect at P, L above; then draw 
lines from P, L through 2, 3, until they intersect at C, 
as the dotted lines show. Take C as a center, and 
with the radius A, describe the semi-circle A, A, and 
with the same radius transferred to C, Fig. 38, describe 
the arc A, B, as shown. Again, with the same radius, 
set off A, B, A, B on Fig. 37, cutting the semi-circle at 
B, as shown. Now draw through B, on the right, 
parallel with S, 3, cutting at J and F; square over F, 
H and J, K, and join H, C; this gives bevel X, as the 
cut for face of sides, which come together at the angle 
shown at 3. The miters on the edge of stuff are 
parallel with the dotted line, L, 3. This is the acute 
corner of the hopper, and as the edges are worked off 
to the bevel 2, as shown in Fig. 38, the miter must be 
correct. 

Having mastered the details of the acute corner, the 
square corner at S will be next in order The first step 
is to join K, V, which gives the bevel Y, for the cut 
on the face of sides on the ends, which form the square 
corners. The method of obtaining these lines is the 
same as that explained for obtaining them for the 
acute-angled corner, as shown by the dotted lines, 
Fig. 35 * A s the angles, S, T, are both square, being 
right and left, the same operation answers both, that 
is, the bevel Y does for both corners. 

Coming to the obtuse angle, P, 2, we draw a line 
B, E, on the left, parallel with A, 2, cutting at E, as 
shown by dotted line. Square over at E, cutting 
T, A, 2 at N; join N, C, which will give the bevel W, 
which is the angle of cut for face of sides. The miters 
on edges are found by drawing a line parallel with P, 2. 

In this problem, like Fig. 34, every line necessary 
to the cutting of a hopper after the plan as shown by 


146 


MODERN CARPENTRY 


the boundary lines 2, 3, T, S. is complete and exhaust¬ 
ive, but it must be understood that in actual work the 
spreading out of the sides, as here exhibited, will not 
be necessary, as the -angles will find themselves when 
the work is put together. When the plan of the base— 
which is the small end of the hopper in this case—is 
given, and the slant or inclination of the sides known, 
the rest may be easily obtained. In order to become 
thoroughly conversant with the problem, I would 
advise the workman to have the drawing made on 
cardboard, so as to cut out all the outer lines, in¬ 
cluding the open corners, which form the miters, 
leaving the whole piece loose. Then make slight 
cuts in the back of the cardboard, opposite the lines 
2, 3, S, T, just deep enough to admit of the cardboard 
being bent 'upwards on the cut lines without breaking. 
Then run the knife along the lines, which indicates the 
edges of the hopper sides. This cut must be made on 
the face side of the drawing, so as to admit of the 
edges being turned downwards. After all cuts are 
made raise the sides until the corners come closely 
together, and let the edges fall level, or in such a 
position that the miters come closely together. If the 
lines have been drawn accurately and the cuts made 
on the lines in a proper manner, the work will adjust 
itself nicely, and the sides will have the exact inclina¬ 
tion shown at Fig. 38, and a perfect model of the 
work will be the result. 

This is a very interesting problem, and the working 
out of it, as suggested, cannot but afford both profit 
and pleasure to the young workman. 

From what has preceded, it must be evident to the 
workman that the lines giving proper angles and 
bevels for the corner post of a hopper must of neces- 


JOINER’S WORK 


14* 

sity give the proper lines for the corner post for a pyr¬ 
amidal building, such as a railway tank frame, or any 
similar structure. True, the position of the post is 
inverted, as in the hopper, its top falls outward, while 
in the timber structure the top inclines inward; but this 
makes no difference in the theory, all the operator has 
to bear in mind is that the hopper in this case is reversed 
—inverted. Once the proper shape of the corner post 
has been obtained, all other bevels can readily be 
found, as the side cuts for joists and braces can be 
•taken from them. A study of these two figures in this 
direction will lead She student up to a correct knowl¬ 
edge of tapered fraong. 


CHAPTER II 


COVERING SOLIDS, CIRCULAR WORK, DOVETAILING AND 

STAIRS 

There are several ways to cover a circular tower roof. 
Some are covered by bending the boarding around 



them, while others have the joints of the covering ver¬ 
tical, or inclined. In either case, the boarding has to 
be cut to shape. In the first instance, where tha joints 

148 





































JOINER’S WORK 


*49 


are horizontal, the covering must be curved on both 

edges. 

At Fig. 39 I show a part plan, elevation, and develop¬ 
ment of a conical tower roof. ABC shows half the 
plan; DO and EO show the inclination and height of 
the tower, while EH and El show the 
development of the lower course of 
covering. This is obtained by using 
O as a center, with OE as radius, and 
striking the curve El, which is the 
lower edge of the board, and corre¬ 
sponds to DE in the elevation. From 
the same center O, with radius OF, 
describe the curve FH, which is the 
joint GF on the elevation. The board, 

EFHI, may be any convenient width, 
as may also the other boards used for 
covering, but whatever the width de¬ 
cided upon, that same width must be 
continued throughout that course. 

The remaining tiers of covering must 
be obtained in the same way. The 
joints are radial lines from the center 
O. Any convenient length of stuff 
over the distance of three ribs, or raft¬ 
ers, will answer. This solution is ap¬ 
plicable to many kinds of work. The 
rafters in this case are simply straight scantlings; the 
bevels for feet and points may be obtained from the 
diagram. The shape of a “gore,” ( when such is 
required, is shown at Fig. 40, IJK showing the base, 
and L the top or apex. The method of getting it out 
will be easily understood by examining the diagram. 
When “gores” are used forcovering it will be necessary 












MODERN CARPENTRY 


150 

to have cross-ribs nailed in between the rafters, and 
these must be cut to the sweep of the circle, where 
they are nailed in, so that a rib placed in half way up 
will require only to be half the diameter of the base, 
and the other ribs must be cut accordingly. 

To cover a domical roof with horizontal boarding we 
proceed in the manner shown in Fig. 41, where ABC 



is a vertical section through the axis of a circular 
dome, and it is required to cover this dome hori¬ 
zontally. Bisect, the base in the point D, and draw 
DBE perpendicular to AC, cutting the circumference 
in B. Now divide the arc, BC, into equal parts, so 
that each part will be rather less than the width of 
a board, and join the points of division by straight 
lines, which will form an inscribed polygon of so many 
sides; and through these points draw lines parallel 










JOINER’S WORK 


151 


the base AC, meeting the opposite sides of the circum¬ 
ference. The trapezoids formed by the sides of the 
polygon and the horizontal lines may then be regarded 
as the sections of so many frustrums of cones; whence 
results the following mode of procedure: Produce, 
until they meet the line DE, the lines FG, etc., form¬ 
ing the sides of the polygon. Then to describe a 
board which corresponds to the surface of one of the 
zones, as FG, of which the trapezoid is a section from 



C 


the point E, where the line FG produced meets DE, 
with the radii EF, EG describe two arcs and cut off 
the end of the board K on the line of a radius EK. 
The other boards are described in the same manner. 

There are many other solids, some of which it is 
possible the workman may be called upon to cover, 
but as space will not admit of us discussing them all, 
we will illustrate one example, which includes within 
itself the principles by which almost any other solid 





































MODERN CARPENTRY 


* 5 * 


may be dealt with. Let us suppose a tower, having a 
domical roof, rising from another roof having an incli¬ 
nation as shown at BC, Fig. 42, and we wish to board 

it with the joints of 
the boards on the 
same inclination as 
that of the roof 
through which the 
tower rises. To 
accomplish this, let 
A, B, C, D, Fig. 42, 
be the seat of the 
generating section; 
from A draw AG 
perpe ndicular to 
AB, and produce 
CD to meet it in E; 
on A, E describe the 
semi-circle, and 
transfer its perim¬ 
eter to E, G by 
dividing it into 
equal parts, and 
setting off corre¬ 
sponding divisions 
on E, G. Through 
the divisions of the 
semi-circle draw 
lines at right angles 
to AE, producing 
them to meet the 
lines A, D and B, C in z, k , /, m , etc. Through the 
divisions on E, G, draw lines perpendicular to them; 
then through the intersections of the ordinates of* the 













































































JOINER’S WORK 


153 


semi-circle, with the line AD draw the lines i, a , k , z , 
/, jj/, etc., parallel to AG, and where these intersect the 
perpendiculars from EG, in points a , z, y , 

«, etc., trace a curved line, GD, and draw parallel 
to it the curved line HC; then will DC, HG be the 
development of the covering required. 

Almost any description of dome, cone, ogee or 
other solid may be developed, or so dealt with under 
the principle as 
shown in the 
foregoing, that 
the workman, it 
is hoped, will ex- 
perience but 
little difficulty in 
laying out lines 
for cutting mate¬ 
rial to cover any 
form of curved 
roof he may be 
confronted with. 

Another class 
of covering is 
that of making 
soffits for splayed doors or windows having circular or 
segmental heads, such as shown in Fig. 43, which exhib¬ 
its a door with a circular head and splayed jambs. 
The head or soffit is also splayed and is paneled as 
shown. In order to obtain the curved soffit, to show 
the same splay or angle, from the vertical lines of the 
door, proceed as follows: Lay out the width of the 
doorway, showing the splay of the jambs, as at C, B and 
L, P; extend the angle lines, as shown by the dotted 
lines, to A, which gives A, B as the radius of the 







154 


MODERN CARPENTRY 


inside curve, and A, C as radius of the outside curve. 
These radii correspond to the radii A, B and A, C in 
Fig. 43; the figure showing the flat plan of the pan¬ 
eled soffit complete. To find the development, Fig. 
43, get the stretch-out of the quarter circle 2 and 3, 
shown in the elevation at the top of the doorway, and 



make 2, 3 and 3B, Fig. 43, equal to it, and the rest of 
the work is very simple. 

If the soffit is to be laid off into panels, as shown at 
Fig. 44, it is best to prepare a veneer, having its edges 
curved similar to those of Fig. 43, making the veneer 
of some flexible wood, such as basswood, elm or the 
like, that will easily bend over a form, such as is 
shown at Fig. 44. The shape of this form is a portion 
of a cone, the circle L being less in diameter than the 








JOINER’S WORK 


*55 


circle P. The whole is covered with staves, which, of 
course, will be tapered to meet the situation. The 
veneer, x, x, etc., Fig. 43, may then be bent over the 
form and finished to suit the conditions. If the 
mouldings used in the panel work are bolection mould¬ 
ings, they cannot be planted in place until after the 
veneer is taken off the form. 

This method of dealing with splayed work is appli¬ 
cable to windows as well as doors, to circular pews in 


A 



churches and many other places where splayed work 
is required. 

A simple method of finding the veneer for a soffit of 
the form shown in Fig. 43 is shown at Fig. 45. The 
splay is seen at C, from which a line is drawn on the 
angle of the splay to B through which the vertical line 
A passes. B forms the center from which the veneer 











i 5 6 


MODERN CARPENTRY 


is descrioed. A is the center of the circular head, for 
both inside and outside curves, as shown at D. The 
radial lines centering at B show how to kerf the stuff 
when necessary for bending. The line E is at right 
angles with the line CB, and the veneer CE is the 
proper length to run half way around the soffit. The 
joints are radial lines just as shown. 


A method for ob¬ 
taining the correct 
shape of a veneer 
for a gothic splayed 
window or door- 
head, is shown at 
Fig. 46; E shows 
the sill, and line 
BA the angle of 
splay. BC shows 
the outside of the 
splay; erect the in¬ 
side line F to A, 
and this point will 
form the center 
from which to de- 
cribe the curve or 
veneer G. This 



Fig. 47 . 


veneer will be the proper shape to bend in the soffit 
on either side of the window head. 

The art of dovetailing is almost obsolete among 
carpenters, as most of this kind of work is now done 
by cabinet-makers, or by a few special workmen in 
the factories. It will be well, however, to preserve the 
art, and every young workman should riot rest until he 
can do a good job of work in dovetailing; he will not 
find it a difficult operation. 


JOINER’S WORK 


157 


There are three kinds of dovetailing, i.e., the com¬ 
mon dovetail, Fig. 47; the lapped dovetail, Fig. 48, 
and the secret, or mitered dovetail, Fig. 49. These 
may be subdivided into other kinds of dovetailing, 
but there will be but little difference. 

The common dovetail is the strongest, but shows the 
ends of the dovetails on both faces of the angles, 



and /i^ therefore, only used in such places as that 
of a drawer, where the external angle is not 
seen. 

The lapped dovetail, where the ends of the dovetails 
show on one side of the angle only, is used in such 
places as the front of a drawer, the side being only 
seen when opened. 

In the miter or secret dovetail, the dovetails are not 
seen at all. It is the weak'est of the three kinds. 





MODERN CARPENTRY 


158 

At Figs. 50 and 51 I show two methods of dovetail¬ 
ing hoppers, trays and other splayed work. The 
reference letters A and B show that when the work is 
together A will stand directly over B. Care must be 



taken when preparing the ends of stuff for dovetailing 
for hoppers, trays, etc., that the right bevels and 
angles are obtained, according to the rules explained 



for finding the cuts and bevels for hoppers and work 
of a similar kind, in the examples given previously. 
All stuff for hopper work intending to be dovetailed 














JOINER’S WORK 


*39 

must be prepared with butt joints before the dovetails 
are laid out. Joints of this kind may be made com¬ 
mon, lapped or mitered. In making the latter much 
skill and labor will be required. 

Stair building and handrailing combined is a science 
in itself, and one that taxes the best skill in the mar¬ 
ket, and it will be impossible for me to do more than 
touch the subject, and that in such a manner as to 
enable the workman to lay out an ordinary straight 
flight of stairs. For further instructions in stair 
building I would refer my readers to some one or 
two of the many works on the subject that can be 
obtained from any dealer in mechanical or scientific 
books. 

The first thing the stair builder has to ascertain is the 
dimension of the space the stairs are to occupy; then 
he must get the height, or the risers, and the width of 
the treads, and, as architects generally draw the plan 
of the stairs, showing the space they are to occupy 
and the number of treads, the stair builder has only to 
measure the height from floor to floor and divide by 
the number of risers and the distance from first to last 
riser, and divide by the number of treads. (This 
refers only to straight stairs.) Let us take an exam¬ 
ple: Say that we have ten feet of height and fifteen 
feet ten inches of run, and we have nineteen treads; 
thus fifteen feet ten inches divided by nineteen gives 
us ten inches for the width of the tread, and we have 
ten feet rise divided by twenty (observe here that 
there is always one more riser than tread), which gives 
us six inches for the height of the riser. The pitch- 
board must now be made, and as all the work has 
to be set out from it, care must be taken to make it 
exactly right. Take a piece of board, same as shown 


i6o 


MODERN CARPENTRY 


in Fig. 52, about half an inch thick, dress it and square 
the side and end, A, B, C; set off the height of the 
rise from A to B, and the width of the tread from B 
toC; now cut the line AC, and the pitch-board is com¬ 
plete, as shown in Fig. 53. This may be done by the 
steel square as shown at Fig. 54. To get the width of 
string-boards draw the line AB, Fig. 53; add to the length 
of this line about half an inch more at A, the margin 
to be allowed, and the total will be the width of 
string-boards. Thus, say that we allow three inches 



for margin, one-half inch to be left on the under side 
of string-board, will make the width of string-boards 
in this case about nine inches. Now get a plank, say 
one and a half inches, of any thickness that may be 
agreed upon, the length may be obtained by multiply¬ 
ing the longest side of the pitch-boards, AC, Fig. 52, 
by the number of risers; but as this is the only class of 
stairs that the length of string-boards can be obtained 
in this way I would recommend the beginner to prac¬ 
tice the sure plan of taking the pitch-board and apply¬ 
ing it as at I, 2, 3, 19, Fig. 55. Drawing all the steps 














JOINER’S WORK 


161 

this way will prevent a mistake that sometimes occurs, 
viz. the string-boards being cut too short. Cut the 
foot at the line AB, and the top, as at CD. This will 
give about one and a half inches more than the 
extreme length. Now cut out the treads and risers; 
the.width of stair is, say, three feet, and we have one 
and a half inches on each side for string-boards. 
Allow three-eights of an inch for housing on each 
side. This will make the length of tread and risers 
two and one-fourth inches less than the full width of 
stairs; and as the treads must project their own thick¬ 
ness over rise, which is, say, one and a half inches, the 
full size of tread will be two feet by eleven and one- 
half inches, and of the risers two feet nine and three- 
fourths inches by six inches; and observe that the first 
riser will be the thickness of the tread less than the 
others; it will be only four and one-half inches wide. 
The reason of this riser being less than the others is 
because it has a tread thickness extra. 

I will now leave the beginner to prepare all his work. 
Dress the risers on one face and one edge; dress the 
treads on one face and both edges, making them all 
of equal width; gauge the ends and the face edge to 
the required thickness, and round off the nosings; 
dress the string-boards to one face and edge to match 
each other. 

A plan of a stair having 13 risers and three winders 
below is shown at Fig. 56. This shows how the whole 
stair may be laid out. It is inclosed between two 
walls. 

The beginner in stair-work had better resort to the 
old method of using a story-rod for getting the num¬ 
ber of risers. Take a rod and mark on it the exact 
height from top of lower flo'or to top of next floor, then 


i6a 


MODERN CARPENTRY 


divide up and mark off the number of risers required. 
There is always one more riser than tread in every 
flight of stairs. The first riser must be cut the thick¬ 
ness of the tread less than the others. 

When there are winders, special treatment will be 



required, as shown in Fig. 56, for the treads, but the 
riser must always be the same width for each separate 
flight. 

When the stair is straight and without winders, a 
rod may be used for laying off the steps. The width 
of the steps, or treads, will be governed somewhat by 
the space allotted for the run of the stairs. 

There is a certain proportion existing between the 
tread and riser of a stair, that should be kept to as close 
as possible when laying out the work Architects 































JOINER’S WORK 


163 


say that the exact measurement for a tread and riser 
should be sixteen inches, or thereabouts. That is, if a 
riser is made six inches, the tread should be ten inches 
wide, and so on. I give a table herewith, showing the 
rule generally made use of by stair builders for deter¬ 
mining the widths of risers and treads: 


Treads 

'Risers 

Treads 

Risers 

Inches 

Inches 

Inches 

Inches 

5 

9 

12 

5 ^ 

6 

8 j 4 

13 

5 

7 

8 

14 

4 ^ 

8 

7/4 

15 

4 

9 

7 / 

l6 

3 '/* 

10 

ey 

17 

3 

11 

6 

18 

2 ^ 


It is seldom, however that the proportion of the 



riser and step is exactly a matter of choice—the room 
















MODERN CARPENTRY 


allotted to the stairs usually determines this propor¬ 
tion; but the above will be found a useful standard, to 
which it is desirable to approximate. 

In better class buildings the number of steps is con¬ 
sidered in the plan, which it is the business of the 
architect to arrange, and in such cases the height 
of the story-rod is simply divided to the number 
required. \ 

An elevation of a stair with winders is shown at 
Fig. 57, where the story-rod is in evidence with the 
number of risers figured o£L 



Fig. 58 shows a portion of an open string stair, with 
a part of the rail laid on it at AB, CD, and the newel 
cap with the projection at A. This shows how the 
cap should stand over the lower step. 

Fig. 59 shows the manner of constructing the step; 
S represents the string, R the risers, T the tread, O 
the nosing and cove moulding, and B is a block glued 
or otherwise fastened to both riser and tread to render 























JOINER’S WORK 


i6 5 

them strong and firm. It will be seen the riser is let 
into the tread, and has a shoulder on the inside. The 
bottom of the riser is nailed to the back of the next 
lower tread, which binds the whole lower part to¬ 
gether. The 
nosing of the 
stair is gen- 
e r a 1 ly re¬ 
turned at the 
open end of 
the tread, 
and this cov¬ 
ers the end 
wood of the 
tread and the 
joints of the 
balusters, as 
shown at 
Fig 60. 

When a stair is bracketed, as shown at B, Fig. 60, 
the point of the riser on its string end should be left 

standing past the string 
the thickness of the 
bracket, and the end of 
the bracket miters 
against it, thus avoid¬ 
ing the necessity of 
showing end wood or 
joint The cove should 
finish inside the length 
of the bracket, and the 
nosing should fin¬ 
ish just outside the 
length of the bracket. When brackets are employed 



















166 MODERN CARPENTRY 

they should continue along the cylinder and all 
around the well-hole trimmers, though they may 
be varied to suit conditions when continuously run¬ 
ning on a straight horizontal facia. 


CHAPTER III 


joiner’s work—useful miscellaneous examples 

I am well aware that workmen are always on the 
lookout for details of work, and welcome everything 
in this line that is new. While styles and shapes 
change from year to year, like fashion in women’s 
dress, the principles of construction never change, 
and styles of finish in woodwork that may be in vogue 
to-day, may be old-fashioned and discarded next year, 
therefore it may not be wise to load these pages with 
many examples of finish as made use of to-day. A 
few examples, however, may not be out of place, so I 
close this section by offering a few pages of such 
details as I feel assured will be found useful for a long 
time to come. 

Fig. I is a full page illustration of three exam¬ 
ples of stairs and newels in modern styles. The 
upper one is a colonial stairway with a square newel, 
as shown at A. A baluster is also shown, so Hi at the 
whole may be copied if required. The second exam¬ 
ple shows two newels and balusters, and paneled string 
and spandril AB, also section of paneled work on end 
of short flight. The third shows a plain open stair, 
with baluster and newel, the latter starting from 
first step. 

At Fig. 2, which is also a full page, seven of the 
latest designs for doors are shown. Those marked 
167 


Modern carpentry 


i 63 

















































































































JOINER’S WORK 


169 





























































































































170 


MODERN CARPENTRY 


ABCD are more particularly employed for inside 
work, while F and G may be used on outside work, 
the five-paneled door being the more popular. 

There are ten different illustrations, shown at Fig. 3, 
of various details. The five upper ones show the gen¬ 
eral method of constructing and finishing a window 
frame for weighted sash. The section A shows a part 
of a wall intended for brick veneering, the upper story 
being shingled or clapboarded. 

The position of windows and method of finishing 
bottom of frame, both inside and out, are shown in 
this section, also manner of cutting joists for sill. 
The same method—on a. larger scale—is shown at C, 
only the latter is intended for a balloon frame, which 
is to be boarded and sided on the outside. 

At B another method for cutting joists for sill is 
shown, where the frame is a balloon one. This frame 
is supposed to be boarded inside and out, and grounds 
are planted on for finish, as shown at the base. There 
is also shown a carpet strip, or quarter-round. The 
outside is finished with siding. 

The two smaller sections show foundation walls, 
heights of .stories, position of windows, cornices 
and gutters, and methods of connecting sills to 
joists. 

A number of examples are shown in Fig. 4 that will 
prove useful. One is an oval window with keys. 
This is often employed to light vestibules, back stairs 
or narrow hallways. Another one, without keys, is 
shown on the lower part of the page. There are three 
examples of eyebrow dormers shown. These are 
different in style, and will, of course, require different 
construction. 

The dormer window, shown at the foot of the page, 


JOINER'S WORK 


J 7 * 









































































































































































172 


MODERN CARPENTRY 














































JOINER’S WORK 


i 73 


is designed for a house built in colonial style, but may 
be adapted to other styles. 

The four first examples in Fig. 5 show the sections 
of various parts of a bay window for a balloon frame. 
The manner of constructing the angle is shown, also 
the sill and head of window, the various parts and 
manner of working them being given. A part of the 
section of the top of the window is shown at E, the 
inside finish being purposely left off. At F is shown 
an angle of greater length, which is sometimes the 
case in bay windows. The manner of construction is 
quite simple. The lower portion of the page shows 
some fine examples of turned and carved work. These 
will often be found useful in giving ideas for turned 
work for a variety of purposes. 

Six examples of shingling are shown in Fig. 6. 
The first sketch, A, is intended for a hip, and is a 
fairly good example, and if well done will insure a 
water-tight roof at that point. In laying out the 
shingles for this plan the courses are managed as fol¬ 
lows: No. I is laid all the way out to the line of 
the hip, the edge of the shingle being planed off, so 
that course No. 2, on the adjacent side will line per¬ 
fectly tight down upon it. Next No. 3 is laid and is 
dressed down in the same manner as the first, after 
which No. 4 is brought along the same as No. 2. The 
work proceeds in this manner, first right and then left. 

.In the second sketch, B, the shingles are laid on the 
hip in a way to bring the grain of the shingles more 
nearly parallel with the line of the hip. This method 
overcomes the projection of cross-grained points. 
Anoth er m ethod of shingling hips is shown at C and 
D. In putting on shingles by this method a line is 
snapped four inches from angle of hip on both sides 




174 


MODERN CARPENTRY 














































































































































































JOINER’S WORK 


175 


of the ridge, as indicated by the dotted lines in C, then 
bring the corner of the shingles of each course to the 
line as shown. When all through with the plain shin¬ 
gling, make a pattern to suit, and only cut the top to 
shape, as the bottoms or butts will break joints every 
time, and the hip line will lay square with the hip 
line, as shown at D; thus making a first-class water¬ 
tight job, and one on which the shingles will not curl 
up, and it will have a good appearance as well. 

At E a method is shown for shingling a valley, 
where no tin or metal is employed. The manner of 
doing this work is as follows: First take a strip 4 
inches wide and chamfer it on the edges on the out¬ 
side, so that it will lay down smooth to the sheeting, 
and nail it into the valley. Take a shingle about 4 
inches wide to start with and lay lengthwise of the 
valley, fitting the shingle on each side. The first 
course, which is always double, would then start with 
the narrow shingle, marked B, and carried up the val¬ 
ley, as shown in the sketch. Half way between each 
course lay a shingle, A, about 4 or 5 inches wide, 
as the case requires, chamfering underneath on 
each side, so that the next course will lie smooth 
over it. 

If tin or zinc can be obtained, it is ,better it should 
be laid in the valley, whether this method be adopted 
or not. 

The sketch shown at F is intended to illustrate the 
manner in which a valley should be laid with tin, zinc 
or galvanized iron. The dotted lines show the width 
of the metal, which should never be less than four¬ 
teen inches to insure a tight roof. The shingles 
should lap over as shown, and not less than four 
inches of the valley,) H, should be clear of shingles 



MODERN CARPENTRY 




ft > 



I 


N 



















































































































































JOINER’S WORK 


*77 


in order to insure plenty of space for the water to 
flow during a heavy rain storm. A great deal 
of care should be taken in shingling and finishing a 
valley, as it is always a weak spot in the roof. 













« 




» 




t 











• •’ 


* 



* w 
















* 






















V 









4 







* 







« 


t 































PART IV 

USEFUL TABLES AND MEMORANDA 
FOR BUILDERS 


Table showing quantity of material in every four 
lineal feet of exterior wall in a balloon frame build¬ 
ing, height of wall being given: 


length of 1 

Studs. 

Size of Sills. 

Size of Studs, 

Braces, etc. 

Quantity of 
Rough dumber 

Quantity of 
Inch Boarding. 

Siding in 
sup. feet. 

Tar Paper 
in sup. feet. 

8 

6x 6 

2x4 studs. 

42 

36 

40 

74 

IO 

6x 8 

4x4 braces. 

52 

44 

50 

80 

12 

6x10 

4x4 plates. 

62 

53 

60 

96 

14 

6x10 

1x6 ribbons. 

69 

62 

70 

112 

16 

8x10 


82 

7i 

80 

128 

18 

8x10 

studs. 

87 

80 

90 

144 

20 

8x12 

16 inches from 

98 

88 

IOO 

160 

22 

9x12 

centers. 

109 

97 

no 

176 

24 

10x12 


119 

106 

120 

192 

18 

IOXIO 

2x6 studs. 

122 

80 

90 

144 

20 

10x12 

6x6 braces. 

137 

88 

IOO 

160 

22 

10x12 

4x6 plates. 

145 

97 

no 

176 

24 

12X12 

1x6 ribbons. 

162 

106 

120 

192 

26 

10x14 


169 

114 

130 

208 

28 

10x14 

studs .16 inch centers. 

176 

123 

140 

224 

30 

12x14 


198 

132 

150 

240 


179 












MODERN CARPENTRY 


180 

Table showing amount of lumber in rafters, collar- 
piece and boarding, and number of shingles to four 
lineal feet of roof, measured from eave to eave over 
ridge. Rafters 16-inch centers: 


Width 

of 

House, 

Feet. 

Size of 
Rafters. 

Size of 
Collar- 
piece. 

Quantity of 

Dumber 
in Rafter 
and 
Collar- 
piece. 

Quantity 

of 

Boarding, 

Feet. 

No. of 
Shingles. 

14 

2x4 

2x4 

39 

6l 

560 

16 

2x4 

2x4 

45 

70 

640 

18 

2x4 

2x4 

50 

79 

720 

20 

2x4 

2x4 

56 

88 

800 

22 

2x4 

2x4 

62 

97 

880 

24 

2x4 

2x4 

67 

106 

960 

20 

2x6 

2x6 

84 

88 

800 

22 

2x6 

2x6 

92 

97 

880 

24 

2x6 

2x6 

101 

106 

960 

26 

2x6 

2x6 

109 

115 

1040 

28 

2x6 

2x6 

117 

124 

1120 

30 

2x6 

2x6 

126 

133 

1200 


A proper allowance for waste is included in the 
above. Roof, one-fourth pitch. 


Table showing the requisite sizes of girders and 
joists for warehouses, the span and distances apart 
being given: 


Distance 

apart. 

Span of Girders. 

Joists. 

Remarks. 

6 Feet. 

8 Feet. 

lOiFeet. 

12 Feet. 

Feet. 

Inches. 

Inches. 

Inches. 

Inches. 

Inches. 


10 

8x12 

12X13 

12X16 

14x18 

2 |XIO 

Girders to have a 

12 

9X12 

12x14 

12X18 

16x18 

3 xio 

bearing at each 

14 

10X12 

12X15 

14x18 


3 xi2 

end and joists 6 in. 
























USEFUL TABLES 


x*i 

Table as before, adapted for churches, public halls, 
etc. 


Distance 

Apart. 


Span of Girders. 

- 

Joists. 

Remarks. 

6 Feet. 

8 Feet. 

10 Feet. 

12 Feet. 

Feet. 

Inches. 

Inches. 

Inches. 

Inches. 

Inches. 


12 

6x10 

8x12 

12x14 

12x16 

2x8 


13 

9XII 

9x12 

11x15 

12X17 

2x9 

Bearings of 

14 

6x12 

10X12 

12X15 

IIXl8 

2x9 

girders and 

15 

7X12 

IIXI2 

IIXl6 

12X16 

2 XIO 

joists as 

16 

8x12 

12X12 

12X16 

13x18 

2 XIO 

above. 

17 

8x12 

9XI4 

12x17 

14x18 

2 X12 


IS 

9x12 

10X14 

IIXI8 


2 X12 


19 

9x12 

IIXI4 

12X18 


2 \ X12 

Both tables 

20 

10X12 

12X14 

13x18 


2 \ X12 

are calcu¬ 

21 

10X12 

IIXI5 

14x18 


2\ XI2 

lated for yel¬ 

22 

IIXI2 

12X15 



3 x 12 

low pine. 


11X12 

IIXl6 



3 xi2 

24 

10X13 

12X16 



3 X13 


25 

10X13 

12X17 



3 X13 


26 

10X14 

I2Xl8 



1 x 14 


27 

10X14 

I2Xl8 



D 4 

3 X14 



Table showing quantity of lumber in every four 
lineal feet of partition, studs being placed 16 centers, 
waste included: 


Height of Partition, 
Feet. 

Quantity of Studs 2x4 
Feet. 

If 2x0 

Feet. 

8 

20 

30 

9 

23 

34 

10 

26 

38 

11 

29 

42 

12 

32 

46 

13 

35 

51 

14 

38 

55 

15 

4i 

59 

16 

44' 

64 

































182 


MODERN CARPENTRY 


Lumber Measurement Table 


Length. 

Length. 

A 

tl) 

a 

u 

►4 

Length. 

A 

to 

O 

V 

►4 

jd 

to 

a 

hT 

2x4 

2x6 

2x8 

2X10 

3x6 

3x8 

12 

8 

12 

12 

12 

16 

12 

20 

12 

18 

12 

24 

14 

9 

M 

14 

14 

19 

14 

23 

14 

21 

14 

28 

16 

11 

16 

l6 

l6 

21 

16 

27 

l6 

24 

16 

32 

18 

12 

18 

18 

18 

24 

18 

30 

18 

27 

18 

3b 

20 

13 

20 

20 

20 

27 

20 

33 

20 

30 

20 

40 

22 

15 

22 

22 

22 

29 

22 

37 

22 

33 

22 

44 

24 

16 

24 

24 

24 

32 

24 

40 

24 

36 

24 

48 

26 

17 

26 

26 

26 

35 

26 

43 

26 

39 

26 

52 


3x10 

3x12 

4x4 

4x6 

4x8 

6x6 

12 

30 

12 

3 b 

12 

16 

12 

24 

12 

32 

12 

36 

14 

35 

14 

42 

14 

19 

14 

28 

14 

37 

14 

42 

16 

40 

16 

48 

16 

21 

16 

32 

16 

43 

16 

48 

18 

45 

18 

54 

18 

24 

18 

36 

18 

48 

18* 

54 

20 

50 

20 

60 

20 

27 

20 

40 

20 

53 

20 

60 

22 

55 

22 

66 

22 

29 

22 

44 

22 

59 

22 

66 

24 

60 

24 

72 

24 

32 

24 

48 

24 

b 4 

24 

72 

26 

65 

26 

78 

26 

35 

26 

52 

26 

69 

26 

78 

6x8 

8x8 

8x10 

IOXIO 

10x12 

12x12 

12 

48 

12 

64 

12 

80 

12 

100 

12 

120 

12 

144 

14 

56 

14 

75 

14 

93 

14 

117 

14 

140 

14 

168 

16 

64 

16 

85 

16 

107 

-l6 

133 

16 

160 

16 

192 

18 

72 

18 

96 

18 

120 

18 

150 

18 

180 

18 

216 

20 

80 

20 

107 

20 

133 

20 

167 

20 

200 

20 

240 

22 

88 

22 

117 

22 

147 

22 

183 

22 

220 

22 

264 

24 

96 

24 

128 

24 

160 

24 

200 

24 

240 ' 

24 

288 

26 

104 

26 

139 

26 

173 

26 

217 

26 

260 

26 

312 


Strength of Materials 

Resistance to extension and compression, in pounds per square 
inch section of some materials. 


Name of the 
Material. 

Resistance 
to Extension. 

Resistance 
to Compression 

Tensile Stre th 
in Practice. 

Comp.Strength 
in Practice. 

White pine... 

10,000 

6,000 

2,000 

1,200 

White oak.... 

15,000 

7,500 

3,000 

1,500 

Rock elm. 

16,000 

8,011 

3,200 

1,602 

Wroughtiron 

60,000 

50,000 

12,000 

10,000 

Cast iron. 

20,000 

100,000 

4,000 

20,000 


In practice, from one-fifth to one-sixth of the 
strength is all that should be depended upon 

























































USEFUL TABLES 


183 


Table of Superficial or Flat Measure 

By which the contents in Superficial Feet , of Boards, Plank, Pav¬ 
ing, etc., of any Length and Breadth , can be obtained, by 
multiplying the decimal expressed in the table by the length 
of the board, etc. 


Breadth 

inches. 

Area of a lin¬ 
eal foot. 

Breadth 

Inches. 

Area of a lin¬ 
eal foot. 

Breadth 

inches. 

Area of a lin¬ 
eal foot. 

Breadth 

inches. 

Area of a lin¬ 
eal foot. 

1 

.0208 

3l 

.2708 

61 

.5208 

9i 

.7708 

1 

.0417 

3l 

.2916 

61 

.5416 

9l 

.7917 

1 

.0625 

31 

•3125 

61 

.5625 

91 

.8125 

1 

.0834 

4 

• 3334 

7 

•5833 

10 

.8334 

n 

.1042 

4 i 

.3542 

71 

.6042 

ioi 

.8542 


.125 

4* 

.375 

71 

.625 

10J 

.875 


•1459 

41 

.3958 

71 

.6458 

10$ 

.8959 

2 

. 1667 

5 

.4167 

8 

.6667 

II 

.9167 

2 i 

.1875 

51 

•4375 

81 

.6875 

III 

•9375 

2* 

.2084 

5* 

.4583 

8| 

.7084 

III 

.9583 

21 

.2292 

51 

.4792 

8f 

.7292 

III 

.9792 

3 

•25 

6 

•5 

9 

•75 

12 

1.0000 


Round and Equal-Sided Timber Measure 

Table for ascertaining the number of Cubical Feet, or solid con- 
^ tents, in a Stick of Round or Equal-Sided Timber, Tree, etc. 


X girt 
in in. 

Area in 
feet. 

He'* 

in in. 

Area in 
feet. 

&g'rt 
in in. 

Area in 
feet. 

M g'rt 
in in. 

Area in 
feet. 

X girt 
in in. 

| Area in 
| feet. 

6 

.25 

iof 


.803 

151 

1.668 

20\ 

2.898 

25 

4-34 

61 

.272 

II 


.84 

155 

1.722 

2ol 

2.917 

251 

4.428 

61 

.294 

Hi 


.878 

16 

1-777 

20f 

2-99 

251 

4.516 

61 

.317 

Hi 


.918 

16I 

1-833 

21 

3.062 

251 

4-605 

7 

•34 

Hi 


•959 

16* 

I.89 

211 

3.136 

26 

4.694 

71 

.364 

12 

1. 

i6§ 

I.948 

211 

3.209 

261 

4.785 

71 

•39 

I2l 

1.042 

17 

2.006 

2l| 

3-285 

261 

4.876 

71 

.417 

121 

in 

00 

O 

M 

171 

2.066 

22 

3.362 

26f 

4.969 

8 

•444 

I2§ 

1.129 

I7l 

2. 126 

22\ 

3.438 

27 

5.062 

81 

.472 

13 

1.174 

171 

2.187 

22\ 

3-516 

271 

5.158 

81 

.501 

I3l 

1.219 

18 

2.25 

22f 

3.598 

27! 

5.252 

81 

• 531 

I3| 

I.265 

18I 

2.313 

23 

3.673 

271 

5.348 

9 

.562 

13} 

I.3I3 

18I 

2.376 

23l 

3.754 

28 

5-444 

9l 

•594 

14 

I.361 

181 

2.442 

23t 

3.835 

281 

5.542 

9l 

.626 

I4i 

I.41 

19 

2. 506 

231 

3 917 

281 

5.64 

91 

•659 

I4i 

1.46 

I9l 

2.574 

24 

4- 

281 

5-74 

10 

.694 

I4f 

I.5H 

I9l 

2.64 

24J 

4.084 

29 

5.84 

i°l 

•73 

15 

I. 562 

19I 

2.709 

24I 

4.168 

291 

5 941 

IQ 1 

.766 

15 i 

1.615 

20 

2-777 

24! 

4.254 

29l 

6.044 

























MODERN CARPENTRY 


184 


Shingling 

To find the number of shingles required to cover 100 
square feet deduct 3 inches from the length, divide 
the remainder by 3, the result will be the exposed 
length of a shingle; multiplying this with the average 
width of a shingle, the product will be the exposed 
area. Dividing 14,400, the number of square inches 
in a square, by the exposed area of a shingle will give 
the number required to cover 100 square feet of roof. 

In estimating the number of shingles required, an 
allowance should always be made for waste. 

Estimates on cost of shingle roofs are usually given 
per 1,000 shingles. 

Table for Estimating Shingles 


Length of 
Shingles. 

Exposure to 
Weather, 
Inches. 

No. of Sq. Ft. of Roof Cov¬ 
ered by 1000 Shingles. 

No. of Shingles Required 
for 100 Sq. Ft. of Roof. 

4 In. Wide. 

C In. Widq, 

4 In. Wide. | 

| 6 In. Wide. 

15 in. 

4 

Ill 

167 

900 

600 

18 

5 

139 

208 

720 

480 

21 

6 

167 

250 

600 

400 

24 

7 

IQ 4 

29I 

514 

343 

27 

8 

222 

333 

450 

300 


Siding, Flooring, and Laths 

One-fifth more siding and flooring is needed than 
the number of square feet of surface to be covered, 
because of the lap in the siding matching. 

1,000 laths will cover 70 yards of surface, and 11 
pounds of lath nails will nail them on. Eight bushels of 
good lime, 16 bushels of sand, and 1 bushel of hair, 
will make enough good mortar to plaster 100 square 
yards. 

Excavations 

Excavations are measured by the yard (27 cubic feet) 
and irregular depths or surfaces are generally averaged 
;n practice. 
















USEFUL TABLES 


i8 S 


Number of Nails Required in Carpentry Work 

To case and hang one door, I pound. 

To case and hang one window, Y pound. 

Base, ioo lineal feet, i pound. 

To put on rafters, joists, etc., 3 pounds to 1,000 feet. 

To put up studding, same. 

To lay a 6-inch pine floor, 15 pounds to 1,000 feet. 

Sizes of Boxes for Different Measures 

A box 24 inches long by 16 inches wide, and 
28 inches deep will contain a barrel, or 3 bushels. 

A box 24 inches long by 16 inches wide, and 
14 inches deep will contain half a barrel. 

A box 16 inches square and 8f inches deep, will 
contain 1 bushel. 

A box 16 inches by 8f inches wide and 8 inches 
deep, will contain half a bushel. 

A box 8 inches by 8f inches square and 8 inches 
deep, will contain 1 peck. 

A box 8 inches by 8 inches square and 4J inches 
deep, will contain 1 gallon. 

A box 8 inches by 4 inches square and 4f inches 
deep, will contain half a gallon. 

A box 4 inches by 4 inches square and 4^ inches 
deep, will contain 1 quart. 

A box 4 feet long, 3 feet 5 inches wide, and 2 feet 
8 inches deep, will contain I ton of coal. 

Masonry 

Stone masonry is measured by two systems, quarry- 
man’s and mason’s measurements. 


MODERN CARPENTRY 


186 

By the quarryman’s measurements the actual con¬ 
tents are measured—that is, all openings are taken 
out and all corners are measured single. 

By the mason’s measurements, corners and piers are 
doubled, and no allowance made for openings less 
than 3' o"x5' o" and only half the amount of openings 
larger than 3' o"x5' o". 

Range work and cut work is measured superficially 
and in addition to wall measurement. 

An average of six bushels of sand and cement per 
perch of rubble masonry. 

Stone walls are measured by the perch (24^ cubic 
feet, or by the cord of 128 feet). Openings less than 
3 feet wide are counted solid; over 3 feet deducted, 
but 18 inches are added to the running measure for 
each jamb built. 

Arches are counted solid from their spring. Corners 
of buildings are measured twice. Pillars less than 
3 feet are counted on 3 sides as lineal, multiplied by 
fourth side and depth. 

It is customary to measure all foundation and dimen¬ 
sion stone by the cubic foot. Water tables and base 
courses by lineal feet. All sills and lintels or ashlar 
by superficial feet, and no wall less than 18 inches 
thick. 

The height of brick or stone piers should not exceed 
12 times their thickness at the base. 

Masonry is usually measured by the perch (contain- 
ing 24.75 cubic feet), but in practice 25 cubic feet are 
considered a perch of masonry. 

Concreting is usually measured by the cubic yard 
(27 cubic feet). 


USEFUL TABLES 


187 

A cord of stone, 3 bushels of lime and a cubic yard 
of sand, will lay 100 cubic feet of wall. 

Cement, 1 bushel, and sand, 2 bushels, will cover 
y/2 square yards 1 inch thick, 4^ square yards y inch 
thick, and 6 % square yards y 2 inch thick; 1 bushel of 
cement and 1 of sand will cover 2^ square yards 
1 inch thick, 3 square yards % inch thick and 
4square yards y 2 inch thick. 

Brick Work 

Brick work is generally measured by 1,000 bricks 
laid in the wall. In consequence of variations in size 
of bricks, no ruie for volume of laid brick can be 
exact. The following scale is, however, a fair average? 

7 com. bricks to a super, ft. 4 in. wall. 

14 *• “ “ “ “ 9 “ “ 

21 “ “ “ “ “ I3 “ *< 

28 “ “ “ “ “ 18 “ “ 

35 “ “ “ “ “ 22 “ “ 

Corners are not measured twice, as in stone work. 
Openings over 2 feet square are deducted. Arches are 
counted from the spring. Fancy work counted 1 y 2 
bricks for 1. Pillars are measured on their face only. 

A cubic yard of mortar requires 1 cubic yard of sand 
and 9 bushels of lime, and will fill 30 hods. 

One thousand bricks closely stacked occupy about 
56 cubic feet. 

One thousand old bricks, cleaned and loosely* 
stacked, occupy about 72 cubic feet. 

One superficial foot of gauged arches requires 
10 bricks. 

Pavements, according to size of bricks, take 38 brick 
on flat and 60 brick on edge per square yard, on an 
average. 


MODERN CARPENTRY 


188 

Five courses of brick will lay I foot in height on a 
chimney, 6 bricks in a course will make a flue 4 inches 
wide and 12 inches long, and 8 bricks in a course will 
make a flue 8 inches wide and 16 inches long. 


Slating 

A square of slate or slating is 100 superficial feet. 

In measuring, the width of eaves is allowed at the 
widest part. Hips, valleys and cuttings are to be 
measured lineal, and 6 inches extra is allowed. 

The thickness of slates required is from y 3 ^ to y 5 y of 
an inch, and their weight varies when lapped from f 
to 6 % pounds per square foot. 

The “laps” of slates vary from 2 to 4 inches, the 
standard assumed to be 3 inches. 


To Compute the Number of Slate9 of a Given 
Size Required per Square 

Subtract 3 inches from the length of the slate, mul¬ 
tiply the remainder by the width and divide by 2. 
Divide 14,400 by the number so found and the result 
will be the number of slates required. 

Table showing number of slates and pounds of nails 
required to cover 100 square feet of roof. 


Sizes of Slate 

Length of Exposure. 

No. Required. 

Nails Required. 

14 in. 

X 

28 in. 

124 in. 

83 

.6 lbs. 

12 

X 

24 

io£ 

114 

.833 

II 

X 

22 

9 * 

138 

I. 

10 

X 

20 

a* 

165 

1-33 

9 

X 

18 

n 

214 

i -5 

8 

X 

16 

6* 

277 

2. 

7 

X 

14 

5 ^ 

377 

2.66 

6 

X 

12 


533 

3-8 











USEFUL TABLES 


189 


Approximate Weight of Materials for Roofs 


Material. 


Average 
Weight, Lb. 
per Sq. Ft. 


Corrugated galvanized iron, No. 20, unboarded. 

Copper, 16 oz. standing seam. 

Felt and asphalt, without sheathing. 

Glass, in. thick... 

Hemlock sheathing, 1 in. thick. 

Lead, about % in. thick... 

Lath-and-plaster ceiling (ordinary). 

Mackite, 1 in. thick, with plaster. 

Neponset roofing felt, 2 layers. 

Spruce sheathing, 1 in. thick. 

Slate, j 3 g in. thick, 3 in. double lap. 

Slate, in. thick, 3 in. double lap. 

Shingles, 6 in. x 18 in., % to weather. 

Skylight of glass, r 3 g to y 2 in-, including frame. 

Slag roof, 4-ply. 

Terne Plate, IC, without sheathing.. 

Terne Plate, IX, without sheathing... 

Tiles (plain), 10% in. x 6 y x % in.—5^ in. to weather. 
Tiles (Spanish) 14^ in. x 10% in.—7^ in. to weather.. 

White-pine sheathing, 1 in. thick. 

Yellow-pine sheathing, 1 in. thick.. 



2 


2 

6 to 8 
6 to 8 


10 


X 

2^ 


6 % 

4/4 


2 


4 to 10 


4 



8 X 

2 }4 

4 


Snow and Wind Loads 

Data in regard to snow and wind loads are necessary 
in connection with the design of roof trusses. 

Snow Load.—When the slope of a roof is over 
12 inches rise per foot of horizontal run, a snow and 
accidental load of 8 pounds per square foot is ample. 
When the slope is under 12 inches rise per foot of run, 
a snow and accidental load of 12 pounds per square 
foot should be used. The snow load acts vertically, 
and therefore should be added to the dead load in 
designing roof trusses. The snow load may be 
neglected when a high wind pressure has been consid¬ 
ered, as a great wind storm would very likely remove 
all the snow from the roof. 

























MODERN CARPENTRY 


190 

Wind Load.—The wind is considered as blowing in 
a horizontal direction, but the resulting pressure upon 
the roof is always taken normal (at right angles) to 
the slope. The wind pressure against a vertical plane 
depends on the velocity of the wind, and, as ascer¬ 
tained by the United States Signal Service at Mount 
Washington, N. H., is as follows: 


Velocity. 

(Mi. per Hr.) 

IO . 

Pressure . 

(I<b. per Sq. Ft.) 


20. 



, 30 ..... . 



40 . 



50 . 



60. 



80. 



100. 




The wind pressure upon a cylindrical surface is one- 
half that upon a flat surface of the same height and 
width. 

Since the wind is considered as traveling in a hori¬ 
zontal direction, it is evident that the more nearly 
vertical the slope of the roof, the greater will be the 
pressure, and the more nearly horizontal the slope, the 
less will be the pressure. The following table gives 
the pressure exerted upon roofs of different slopes, by 
a wind pressure of 40 pounds per square foot on a 
vertical plane, which is equivalent in intensity to a 
violent hurricane. 


UNITED STATES WEIGHTS AND MEASURES 
Land Measure 

1 sq. acre = 10 sq. chains = 100,000 sq. links = 6,272,640 sq. in. 
1 “ “ = 160 sq. rods = 4,840 sq. yds. = 43,560 sq. ft. 

Note. —208.7103 feet square, or 69.5701 yards square, or 220 feet 
by 198 feet square=i acre. 


















USEFUL TABLES 


191 


Cubic or Solid Measure 

1 cubic yard = 27 cubic feet 
l cubic foot = 1,728 cubic inches. 

1 cubic foot = 2,200 cylindrical inches. 
1 cubic foot = 3,300 spherical inches. 

1 cubic foot = 6,600 conical inches. 


Linear Measure 


12 inches (in.). 

3 feet. 

5.5 yards... 

40 rods. 

8 furlongs .. 

in. ft. 

36= 3 

198 = 16.5 

7,920 = 660 

03,360 = 5,280 


= 1 foot... 

= 1 yard. 

= 1 rod.. 

= 1 furlong. 

= 1 mile. 

yd. rd. fur. mi. 

= 5.5 = 1 

= 220 = 40 = 1 

= 1,760 =320 = 8 = 1 


.ft. 

. y<*- 

.rd. 

.fur. 

.....mi. 


Square Measure 

144 square inches (sq. in.) = 1 square foot...sq, ft. 

9 square feet.= 1 square yard.sq. yd. 

301 square yards.= 1 square rod.sq. rd. 

160 square rods.= 1 acre.A. 

640 acres.= 1 square mile.sq. mi. 

Sq. mi. A. Sq. rd. Sq. yd. Sq. ft. Sq. in. 

1 = 640 = 102,400 = 3,097,600 = 27,878,400 = 4,014,489,600 


Miscellaneous Measures and Weights 

1 perch of stone = 1 ft. X 1 ft. 6 in. X 16 ft. 6 in. = 24.75 ft. cubic. 
1 cord of wood, clay, etc., = 4 ft. X 4 ft. X 8 ft. = 128 ft. cubic. 

1 chaldron = 36 bushels or 57.25 ft. cubic. 

1 cubic foot of sand, solid, weighs 112^ lbs. 

1 cubic foot of sand, loose, weighs 95 lbs. 

1 cubic foot of earth, loose, weighs 93! lbs. 

1 cubic foot of common soil weighs 124 lbs. 

1 cubic foot of strong soil weighs 127 lbs. 

1 cubic foot of clay weighs 120 to 135 lbs. 

1 cubic foot of clay and stone weighs 160 lbs. 

1 cubic foot of common stone weighs 160 lbs. 

1 cubic foot of brick weighs 95 to 120 lbs. 

1 cubic foot of granite weighs 169 to 180 lbs. 

1 cubic foot of marble weighs 166 to 170 lbs. 

1 cubic yard of sand weighs 3,0^7 lbs. 

1 cubic yard of common soil weighs 3,429 lbs. 





















192 


MODERN CARPENTRY 


Safe Bearing Loads 

Brick and Stone Masonry. 

Tb. per 

Sq. In. 

Brick Work » 

Bri^k^ hard laid in lime mortar... 

IOO 

Ward laid in Portland cement mortar.. 

200 

ffard laid in Rosondale cement mortar. 

150 

700 

350 

350 

175 

80 

•• Masonry. 

Granite capstone .. 

Squared stonework. 

Sandstone capstone . 

Squared stonework. 

Rubble stonework, laid in lime mortar. 

Rubble stonework, laid in cement mortar. 

150 

500 

250 

80 

Limestone capstone. 

Squared stonework. 

Rubble, laid in lime mortar. 

Rubble, laid in cement mortar. 

150 

150 

Concrete, 1 Portland, 2 sand, 5 broken stone. 


Foundation Soils. 

Tons 

per Sq. Ft 

Rock, hardest in native bed. 

100 — 

Equal to best ashlar masonry. 

25-40 

15-20 

4- 6 

Equal to best brick. 

Clay, dry, in thick beds. 

M'oderately dry, in thick beds. 

*T v 

2 4 

Soft. 

I- 2 

Gravel and course sand, well cemented. 

8-10 

Sand, compact and well cemented. 

4- 6 

Clean, dry. 

*r u 

2 - 4 

Quicksand, alluvial soil, etc. 

. 5 - 1 


Capacity of Cisterns for Each io Inches in Depth 


Twenty-five feet in diameter holds.3059 gallons 

Twenty feet in diameter holds.1958 gallons 

Fifteen feet in diameter holds.1101 gallons 

Fourteen feet in diameter holds. 959 gallons 

Thirteen feet in diameter holds.;. 827 gallons 

Twelve feet in diameter holds. 705 gallons 

Eleven feet in diameter holds. 592 gallons 

Ten feet in diameter holds. 489 gallons 

Nine feet in diameter holds.... 396 gallons 

Eight feet in diameter holds. 313 gallons 

Seven feet in diameter holds. 239 gallons 

Six and one-half feet in diameter holds. 206 gallons 

Six feet in diameter holds. 176 gallons 

Five feet in diameter holds. 122 gallons 

Four and one-half feet in diameter holds. 99 gallons 



















































USEFUL TABLES 


>93 


Four feet in diameter holds. 78 gallons 

Three feet in diameter holds. 44 gallons 

Two and one-half feet in diameter holds. 30 gallons 

Two feet in diameter holds. 19 gallons 


Number of Nails and Tacks per P^und 



nails. 

No. 


TACKS. 

No. 

Name. 

Size. 

per 

lb. 

Name. 

Length. 

per lb. 

3 penny, fine 1 % inch 760 nails 

I OZ... 

....H inch.... 


3 

‘ . iH 44 

480 


1)4 "... 



4 

‘ . i)4 “ 

300 


2 “... 



5 

.i* “ 

200 


2 % "... 

....5-16 “ .... 


6 

4 .23* “ 

160 


3 

....H " .... 

■••• 5,333 

7 

. 2'/ “ 

128 

44 

4 "... 



8 

4 .2^ “ 

92 


6 “... 


.... 2,666 

9 

.“ 

72 

44 

8 “... 

—•# 44 .... 


10 

4 .3 

60 

4< 

10 “... 



12 

.3 H “ 

44 


12 “... 

“ .... 

... 1,333 

16 

4 .3 K “ 

32 

44 

14 "... 


.... 1,143 

20 

- .4 

24 

44 

16 “... 

....% " .... 


30 

. 4'X “ 

18 

44 

18 "... 


.... 888 

40 

4 . 5 

14 

4 4 

20 “... 



50 

4 . s'A 44 

12 

44 

22 “ ... 



6 

‘ fence 2 “ 

80 

4 4 

24 “... 

....iM “ .... 


8 

‘ “ 2)4 44 

50 





10 

4 44 g 44 

34 

4 4 




12 

4 44 3% " 

29 

4 4 


. 



Wind Pressures on Roofs 

(Pounds per Square Foot.) 


Rise, Inches per 
Foot of Run. 

Angle with 
Horizontal. 

Pitch. 

Proportion of 

Rise to Span 

Wind Pressure, 
Normal to Slope. 

4 

18 0 25' 

6 

16.8 

6 

26° 33' 

i 

23.7 

8 

33 ° 4 i' 

i 

29.I 

12 

45 '* 0* 

l 

36.1 

16 

53 ° 7 ' 

1 

38.7 

18 

56° 20' 

5 

39-3 

24 

63° 27' 

1 

40.0 


In addition to wind and snow loads upon roofs, the 
weight of the principals or roof trusses, including the 
other features of the construction, should be figured in 
the estimate. For light roofs, having a span of not 
over 50 feet, and not required to support any ceiling, 
the weight of the steel construction may be taken at 
5 pounds per square foot; for greater spans, 1 pound 
per square foot should be added for each 10 feet 
increase in the span. 




































































SUPPLEMENT TO 

MODERN CARPENTRY AND JOINERY. 


The aim in preparing this has been to supply neces¬ 
sary information for enabling a practical joiner to be¬ 
come a competent airtight-case maker, without the 
tedium of waiting, perhaps for years, until chance brings 
him into contact with one who has made a specialty of 
this class of work. I have endeavored, by means of 
illustrations, to elucidate as clearly as possible the points 
which are so frequently the cause of failure to those 
who, while having a good knowledge of wood-working, 
have not had the advantages of direct practical tuition 
in the intricacies of airtight-case making. 

Before proceeding with the explanations, I would 
point out that the first and most important rule in join¬ 
ery is to have the stuff planed up true, and gauged 
accurately to size. 


i. airtight wall case with glass or wood ends. 

The general drawings of the airtight wall-case with 
glazed ends are given in Figs, i to 5 and the details in 
Figs. 6 to 9. 

Framework. Figs. 6 and 7 give the width of the top 
and bottom rails for the front frame of the case, and, by 
adding the width of the top and bottom door-rails to 
each we determine the width of the rails required for 
the ends of the case, as shown in Fig. 5. The angle- 
stile must be mc h more in thickness than the thick- 
195 


MODERN CARPENTRY 


196 



ness of the doors, in order to allow of a rebate being 
formed to receive the glass at the ends of the case. (See 
M Fig. 8.) 























CASE MAKING 

|E 


197 



In setting out the framework (which is mortised and 
tenoned together in the ordinary way) the face shoul¬ 
ders of the front rails should be inch longer than the 










































198 


MODERN CARPENTRY 


back shoulders. An eighth inch bead—for which the 
allowance has been made—is worked on the angle-stiles 
and bottom rail only, the edge of the top rail being left 
square. The moulding which is planted round the case, 
as shown in Fig. 6, serves to break the joint of the doors. 
The shoulders on the end rails are square with each 
other, the rebate being the same depth as the moulding. 

Airtight joints. To make successfully the airtight 
joint between the angle-stile of the case and the hanging 
stile of the door (see Fig. 8) three planes are required. 
The first plane is used on the angle-stile for forming at 
the same time the two grooves, each 3/16 inch wide; 



the second is used for working the two fillets together 
and the third for working the two hollows in the door 
stiles. 

The front part of the frame must now be fitted to¬ 
gether and the joints at the back of the frame cleared 
off, to allow the airtight planes to be worked from the 
back of the frame, that is, from the inside of the case, 
as the doors would not*close accurately if they were 
worked from the face or outside. 

After the front frame has been fitted together as de¬ 
scribed, it must be taken apart, and the angle-stiles 
worked with plane No. 1. When this has been done, 










CASE MAKING 


199 



the fillets must be glued in the grooves, and, when set, 
rounded over with plane No. 2. The fillets will not 
require to be taken to the exact width before rounding 
over, as plane No. 2 works all surplus stuff away. 






















200 


MODERN CARPENTRY 


For the joint between the top and bottom rails of doors 
and the airtight fillets respectively, two planes are re- 



quired; the first for sticking the airtight fillet, and the 
second for working the small hollow on the door rails to 
match the fillet. 























CASE MAKING 


201 


Continuing with the framework. After rounding the 
fillets in the angle-stiles, groove the top and bottom 



rails to receive the tongue on the airtight fillets as shown 
in Figs. 6 and 7 and rebate the bottom rails to rest on 
the plinth, Fig. 7. The top and bottom rails at each end 
















































202 


MODERN CARPENTRY 


of the case are trenched to receive respectively the ends 
of the inside top and inside bottom, Fig. 5. Care must 
be taken to make these trenches in such a way as to 
keep the inside top and the inside bottom in the positions 
shown in the Figs. 6 and 7. Rebate the back angle-stile 



of each end frame to receive the back, as in Fig. 8, and 
run a small hollow in the angle of the rebate. Glue and 
pin the airtight fillet on the front edges of the inside 
top and bottom respectively; also glue the fillet on the 
back of each in order to strengthen the airtight fillet, and 










CASE MAKING 


203 


make out the thickness to receive the glue-blocks as 
shown. An ovolo or other moulding is now worked on 
the external angles of the two front angle-stiles as shown 
in Fig. 8, the moulding being stopped in a line with the 
top and bottom rails respectively of the doors, Fig. 1. 

The body of the case must now be put together, care 
being taken to glue-block the front frame and ends se¬ 
curely to the bottom and top, as well as behind the plinth, 
which is screwed to the bottom rails from the back. 

Match-boards are used for the back, the boards being 
run to the floor, as shown in Fig. 2. Mitre the cornice 
round the front and ends, screwing it from the back of 
the top rails; cut the dust-board to fit on the top edges 
of the rails and bevel against the cornice; having pre¬ 
viously rebated it to receive the back of the case. Be¬ 
fore the back is fastened, the cloth, Fig. 8, should be 
placed in the rebate of the stile, the fillet placed on top 
of the cloth and pressed into the hollow, and then fas¬ 
tened to the stile with screws, the cloth thus being se¬ 
curely held between the fillet and the stile. The cloth 
can now be stretched taut and fixed at the other end in 
the same way, and the boards fastened in. 

Doors. In planing up the stuff for the doors, the same 
gauge must be used as that for the frame of the case. 
When setting out for the doors, take the width and 
height accurately, and allow 1/16 inch on the height 
for fitting in. The width is set out as for ordinary 
folding doors, viz.: allowing half the hook-joint on each 
door, and inch for jointing and fitting in. The best 
way to allow for fitting is to have each stile 1/16 inch 
greater in width than the finished size required. 

The rails abutting against each angle-stile are single- 
mortised and tenoned together as in ordinary work, 


204 


MODERN CARPENTRY 


but double mortises and tenons must be used at the top 
and bottom of each meeting stile, as shown in Fig. 9. 
The reason for using the double tenon is, that if a sin¬ 
gle tenon were used, the ends of the tenon would slip 
off after the hook-joint had been made. 

Presuming the doors to be wedged up, level off the 
joints at the shoulders, when the doors will be ready 
for jointing together and fitting to the case. 

Hook-Joint. The following is the best method of 
making a well fitting joint. First rebate the stiles (the 
rebate being T /s inch less in width than the thickness of 
the doors, and 5/16 inch deep), and next bevel the edges 
of the doors, bringing the rebate to a depth of inch, 
Fig. 8. The doors must now be worked with a hollow 
and round on the edge of the rebate to form the hook- 
joint. For this purpose a hook-joint plane is required. 
There is an adjustable depth-gauge on the side of the 
plane, which can be easily set for working different thick¬ 
nesses of stuff. Before working the doors with the plane, 
it is advisable to work a piece of stuff of the same thick¬ 
ness as the doors. Cut the piece thus worked into two, 
and put the joint together. This will test the accuracy 
of the setting of the plane. If the faces do not come 
flush with each other, the gauge on the plane must be 
raised or lowered accordingly. 

Having fitted the meeting stiles, place the doors to¬ 
gether across the bench, as they can thus be more easily 
taken to the exact width and height of the frame of the 
case. After the doors have been fitted in the opening, 
work with the airtight planes as previously instructed, 
always remembering to hold the fence of plane No. 3 on 
the back side of the door while forming the hollows on 
the hanging stiles. With plane No. 2 the small hollow 


CASE MAKING 


205 

on the top and bottom rails to match the airtight fillet is 
worked. 

After working the doors as described, clean off the 
back side, place the doors in position, and clean off the 
face to the level of the frame. Take the doors out and 



work the bead on the joint between the doors, Fig. 8. 
This bead is flatter than usual and has a very small 
quirk. 

The doors are hung to the frame, each by three hinges. 
The top and bottom hinges are usually kept their own 













206 


MODERN CARPENTRY 


depth from the top and bottom edges of the doors re¬ 
spectively, e. g., a 2 y 2 inch hinge will be 2J/2 inches from 
the edge. The handles on the meeting stiles are re¬ 
spectively about 9 inches from the upper and lower edges 
of door. 

All glass in the doors must be carefully packed with 
small slips of wood between the edges of the glass and 
the frame of the door, in order to keep the frame rigid. 
The woodwork being so slight, the doors would sag 
when hung if the glass were not packed tightly, as all 
the weight of the glass would fall on the bottom rail. 


DOOR Hanging stile, of frame. DOOR 



Shelves. The following is the best method to adopt 
for fitting the case with shelves, as, when fitted in this 
way, the shelves can be moved to any required height. 
To the back of the case screw two pieces of iron, one at 
each side, extending from the top to the bottom of the 
case. These must previously have been drilled and 
tapped their whole length, the space between each hole 
being y 2 inch from centre to centre, and each hole being 
large enough to receive a 3/16 inch screw. A malleable- 
iron bracket about 3 inches long on the back edge—the 
length of the top edge being the width of the shelf—is 
now required, having a small piece projecting above the 
top edge in which is drilled a plain hole, and having 
a pin near the bottom edge. The pin at the bottom edge 














CASE MAKING 


207 


is placed in one of the holes in the tapped bar, and a 
3/16 inch screw is passed through the hole at the top 
edge and screwed into the bar, thus securing the bracket 
firmly. Care must be taken to have the same distance be¬ 
tween the centres of any two holes in the bar. 

Fig. 10 shows a horizontal section through a show¬ 
case having solid ends. 

Fig. 11 shows a horizontal section through the centre 
hanging stile in the front frame of a wide showcase, 
when two pairs of doors are required. It is worked in 
the same manner as previously described for hanging 
stiles. 



• * * * 0 « 

1—i—1—.—1->— 

SCALE 

Fig. 12. 


3 '!» 


Fig. 12 shows a section of a cross bar in doors. This 
is only required where sheet glass is used. Each end of 
the bar is sunk into the moulding of the door-stiles. The 
saddle is cut between the rebates, and secured to the bar. 

Plinths separate from the case. If the showcase is 
over 6 feet 6 inches in height, or the plinth is of a greater 
depth than 12 inches, it is advisable to make the plinth 
separate from the case. Instead of the bottom rail being 
rebated behind the plinth, as shown in Fig. 7, a frame 
must be made out of 1^ inch by 3 inch stuff dovetailed 
together at the angles; and two or three bearers should 












208 


MODERN CARPENTRY 



CO 

i—l 

do 

W 


be mortised and tenoned between the front and back 
rails (as the length of the case may require). At each 
angle, and under each end of the bearers, a leg is stump- 


































CASE MAKING 


209 


tenoned into the under side of the rails to support the 
case. When this is done, the plinth should be mitred 
round the frame. It should lx screwed from the back, 
and glue-blocks used in all the angles. 



An isometrical projection of a counter-case is shown 
in Fig. 3. The top, sides, and front are of plate-glass. 
Mirrors are placed on the inside of the doors at the back 
of the case. The divisions on the bottom show the posi¬ 
tion of the trays. 



SCALE 


Fig. 15. 

Before commencing work, it is absolutely necessary 
to draw Figs. 14 and 15 full size, to enable the taking 
off and working to an exact size of the various carts 
required to be done. ' 































210 


MODERN CARPENTRY 


Bottom of case. Commence with the frame, which 
should be made out of well-seasoned pine. The width 
of the bottom frame will be the extreme width of the 
case less the thickness of the moulding on the front edge 
and i 1 /* inch for a hardwood slip on the back edge of 
the frame, Fig. 17. The length will be the extreme 
length of the case minus two thicknesses of moulding. 

Mortise and tenon the frame together, and rebate it 
to receive Y inch panels flush on the inside; then glue 
up and take to size. The hardwood slip can now be 
jointed and glued on, a tongued and grooved joint being 
used for the purpose. After this has been done, the air- 



Fig. 16. 


tight rebate to receive the doors should be worked on 
the hardwood slip. In order to make a good job of the 
rebate, it will be necessary to have a special plane for 
working both the rebate and the small half-round tongue 
at one time. 

To complete the bottom, groove the front edge and 
both ends for the tongue, then mitre and fix the mould¬ 
ing to the frame. The moulding must be specially noted. 
It must project above the bottom 3/16 inch to form a 
rebate for the glass; and the first member, i. e. the part 
projecting, must be rounded to intersect with the upright 
angle-bars, Figs. 17 and 18, with mitre into the mould¬ 
ings. 












CASE MAKING 


211 


The panels in the bottom are to be screwed to the 
frame. Before putting the whole case together, they 



must be taken out for enabling the small fillets which se¬ 
cure the glass to be easily screwed into their respective 
positions. ' 































212 


MODERN CARPENTRY 


Framework for glass. Plane up the stuff for the 
round angle-bars, gauging it to 9/16 inch square, and 
rebate *4 inch deep and inch from the face edges. 
The angle bars will then appear as seen in Fig. 2. For 
the back part of the frame, square up the stuff to 1J4 
inch by ^ inch an d rebate % inch deep and inch 
from the face for the glass. For the doors, take out 



fhs 



I-'ig. 18. 


the rebate ^4 inch deep by $/& inch wide; bevel the re¬ 
bate to 5/16 inch deep on the outside edge (as shown 
in Fig. 21), and work the hook-joint plane on the edge 
of the rebate. It is best to make the mitred joints first, 
as they require careful fitting together, and the bottom 
ends can be afterwards easily taken to the required length 
and cut. 

Fig. 23 contains isometrical projections showing the 






















Door stilt 


CASE MAKING 


213 


joints at the intersection of the front and the end angle- 
bars with the upright angle-bar. The position of the 
point is shown at A, Fig. 23. 




Three pieces of the required section, Fig. 20, should 
be got out, and the joint worked as follow?: 

Commence with .the front and end angle-bars, cutting 


orb bottom/ 

































214 


MODERN CARPENTRY 


a square mitre, 45 degrees on each outside face of both 
bars, bringing the external angle to a point, as shown in 
the sketch. Cut the mitre down to the rebate line and 
cut the surplus away, leaving the core of the bar pro¬ 
jecting, which will be the part C. The internal part of 
the mitre E is the sight line. Square down and across 



the core; then, from the sight-line, measure distances 
of inch and 7/16 inch; the resulting lines will be the 
shoulder and end of the dovetail respectively. Cut the 
core off at the longest line and form the dovetail as 
shown in the sketch, when the two bars can be fitted 
together. 

Proceed with the upright angle-bar. Cut the square 

















CASE MAKING 


215 

mitre as before, but instead of cutting to the depth of 
the rebate, it must be cut 1/32 inch less. From the 



sightline F measure the same distances as before, viz., 
% inch and 7/16 inch. Cut,off at the longest line, tak¬ 
ing care not to cut through the projecting point of the 


Fig. 23. 





2l6 


MODERN CARPENTRY 





























CASE MAKING 


217 


mitre, then take out the core C back to the shoulder line, 
thus leaving a thin tenon as seen in the sketch. Cut 
the tenon back 1/16 inch on each edge and continue the 
mitre through. 

It will now be necessary to mortise the front and end 
bars to receive the tenon on the upright angle-bar. For 
the mortises, square a line across the mitre 1/16 inch 
from the sight line E. Gauge a line down the mitre 
3/32 inch from the face of the bar, leaving 1/32 inch 
(the width of the mortise) between the core of the bar 
and the gauge line. The depth of the mortise will be 
to within 34 inch from the other face. 

The work must be done very carefully, and great care 
taken to have the tenon on the upright angle-bar of the 
thickness stated, viz., 1/32 inch, as the result of having 
it of greater thickness would be that, when the bars were 
rounded, it would work through to the face. 

The front angle-bar will have the same joint on both 
ends. The joint at the back of the case on the end 
angle-bar is cut as shown at Fig. 24. The joint at the 
bottom end of each upright angle-bar is simply a square 
shoulder cut to the depth of the rebate, leaving the core 
of the bar projecting to form a stump tenon. The bars 
are afterwards mitred with the moulding on both the 
front and the end, the projecting round of the moulding 
being cut away between the mitres in order to allow the 
shoulder to butt on the first square member, which will 
be flush with the bottom. 

Fig. 24 contains isometrical projections showing the 
joints used to unite the back rail with the back upright 
angle-bar for forming the door opening; and also the 
end angle-bar. The position of the joint will be clearly 
understood by referring to B, Fig. 13. 


218 


MODERN CARPENTRY 


It will be well to follow the same system as in the last 
group of joints, i. e., to prepare a piece of the required 
section of back rail, Fig. 21, which, when cut into two 
parts, can be used for both the back rail and back angle- 
bar; the only difference in the section of the two being 
that the back rail is rebated 1/16 inch less than the 
thickness of the doors instead of l /% inch less as in the back 
upright bar, Fig. 22. The reason for this is to allow the 



round of the hook-joint on the back upright bar to project 
over the hook-joint on the back rail which butts against 
it. It also allows a continuous hollow on the edges of 
the doors, which would not be the case if the rebates 
were kept flush with each other. 

The end angle-bar is dovetailed into the back rail and 
is also mitred both at the extreme end and at the rebate. 
Fig. 25 shows the plan of this joint. It will be observed 










CASE MAKING 


219 


that the joint has been left open to show the bevel from 
the shoulder line to the dovetail on the back rail, as at A, 

Fig- 31- 

The back rail is also dovetailed to receive the upright 
bar. If the reader will look at Fig. 24 and imagine the 
upright placed into position on the back rail, he will no¬ 
tice that D D meet and form the remaining part of the 



mitre, leaving a shoulder and mitre to join the end angle- 
bar when in position. The exact position of the latter is 
seen in Fig. 26, the dotted lines showing the position of 
the dovetail on the back rail. 

We will now proceed to set out the work. 

Commencing with the end angle bar, square off a line 
for the extreme end of the mitre at B, Fig. 25, and 
measure back the width bf the back rail (namely i l / 2 









220 


MODERN CARPENTRY 


inch) to C, which will be the sight line. From the sight 
line set off 5/16 inch for the shoulder of the dovetail as 
at S, Figs. 24 and 25; then set off inch from the 
sight-line to the end of the dovetail. Set a gauge to 
the centre of the angle-bar for the shoulders, as at D, 
Figs. 25 and 26. The shoulder at D, Fig. 25, is cut 
under on the bevel as shown in the section through the 
joint at A, and in the sketch of the end angle-bar, Fig. 
24, where the drawing is broken. It is necessary to 
bevel it in this way in order to obtain the requisite 
strength in the dovetail. The shoulder on the side, Fig. 
26, is cut square, as shown in the sketch. Mark the 
mitres, cutting from the sight-line to the shoulder line. 
The mitre on the extreme end is cut through as shown 
in Fig. 25. 

To set out the back rail as shown in Fig. 24, square 
a line for the extreme end of the mitre, and from this 
line measure back for the sight-line, namely, 9/16 inch, 
the width of the angle-bar, as at E, Fig. 25. Square a 
line between the two lines obtained, at an equal distance 
from each for the shoulder D. From E measure 7/16 
inch toward the end of the bar, and cut off square to 
within Ys inch of the outside edge; this is clearly shown 
in Fig. 24. 

To mark the dovetail of the end angle-bar, make a 
thin hardwood or zinc pattern to fit the dovetail on the 
angle-bar and apply it to the rebate of the back rail, 
cutting the dovetail out very carefully to within Y inch 
of the outside edge. On the top side of the rail mark 
the external mitre from the extreme point to the shoul¬ 
der-line, and cut as shown in Figs. 24 and 25. Before 
the mitre can be completed, the bevel must be cut along 
the shoulder-line and edge of dovetail, and must work 


CASE MAKING 


221 


out against the mitre. The internal mitre is cut from 
the sight-line. 

There now only remains the cutting of the dovetail 
to receive the upright bar. Referring to Fig. 24, it will 
be seen that it is necessary to obtain the shoulder-line 
only, which is accomplished by measuring from the ex¬ 
treme point of the mitre, D, Fig. 24, ^ inch, the thick¬ 
ness of the upright bar. The position of the dovetail- 
joint between the back rail and the back upright bar is 
shown by the dotted line in Fig. 26. 

Exact lines for setting out the back upright bar, Fig. 
26, are found as follows: Square the shoulder-line D 
and set off for the back shoulder 34 inch as shown by 
the dotted line G. The back shoulder is then cut off to 
within 34 inch of the face, as in the sketch, Fig. 24. 
Make a pattern to fit the dovetail on the back rail, and 
apply it to the back of the bar. Mitre the 34 inch pro¬ 
jection on the outside edge, and also mitre the inside as 
shown. 

It is absolutely necessary that the whole of this work 
should be executed very carefully and very neatly. When 
the above mentioned joints have been fitted, make the 
bars to the required length. 

To set out the bottom end of the back upright 
bar, cut the face shoulder square and mitre with the 
moulding as previously described for the front angle-bar. 
Allow the. back-shoulder to be 34 inch longer, so as to 
fit the rebate for the doors, the tenon being in the po¬ 
sition shown by the dotted lines in Fig. 17. 

After all the joints have been made, round the angle- 
bars and the back rail. The external angles of all 
upright angle-bars must have the rounding turned out 
about 34 inch above the bottom shoulder, leaving the 


222 


MODERN CARPENTRY 


bottom part of the bar square to follow the line of the 
moulding. The joints can now be glued together and 
cleaned off. 

The double-rebated upright bar between the doors, 
as at H, Fig. 19, is cut to give both the top and bottom 
rebate, a small dovetail being cut at both ends in the 
positions shown by the dotted lines. The front edge of 
the bar is slightly rounded to break the joint between 
the doors. From the inside of the bar a runner of the 
same thickness as the bar is screwed to the bottom of 
the case to keep the trays in position. 

Doors. There is nothing special to note in framing 
up the doors; they may be either tenoned or dowelled 
together. The panel is prepared flush on the inside. 

Carefully fit the doors to the opening and work the 
hook-joint on the top edge and both ends. It will be 
remembered that the hook-joint must be worked through 
on each end; and also that it is deeper than the hook- 
joint on the top rail. In working the small hollow to 
fit over the fillet on the bottom edge, work the plane 
from the back side of the door. 

Hinge the doors on the bottom edge, fixing the butts 
against the outside edge of the half-round fillet. When 
fixed thus the airtight joint will remain intact. The 
doors are fastened by a spring catch or lock let into 
the top rail. 

When the doors are hung, the position of the mirror 
fillet can be marked by lining down the back of the 
doors round the frame. The fillets should be fixed 1/32 
of an inch inside the lines to allow for clearing. 

Trays. A cross section of the tray is shown in Fig. 
18. The bottom is prepared for three pieces of * 4 -inch 
pine. The grain of the centre piece runs from back to 


CASE MAKING 


223 


front of the case, the grain of the side pieces being at 
right angles to it, and the three pieces are tongued and 
grooved together as shown. Glue the pieces together, 
and, when set, mitre the bead round the bottom. 

Another method of ensuring the bottom against warp¬ 
ing is to have the bottom in three thicknesses, the grain 
of the centre lying across the two outside pieces, and the 
pieces being glued together. 

The inside of the tray and over the bead are covered 
with velvet or some other material, which must be glued 
to the tray. Glue should be used sparingly so as to 
prevent it penetrating the material. 


CIRCULAR-FRONTED COUNTER-CASE WITH GLASS ENDS. 

Fig. 28 shows a cross section through a circular- 
fronted case with glass ends. The only difference in 
the construction of this case from that of the square 
case is the bent angle-bar, and, of course, the omission 
of the front angle-bar. 

In making this case it is first necessary to have the 
glass bent to the shape required. For this purpose a 
pattern of the curve should be sent to a glass manufac¬ 
turer. When the glass has been received make a mould 
of the same shape, on which to bend the angle-bar, as 
shown in Fig. 29. The convex side of the glass will 
give the rebate line from which to work the mould. 

Use birch for the angle-bar, as it bends easily; it can 
be stained to match the other part of the case. Have 
the bar long enough to bend from the bottom of the case 
to the back rail. 

To bend the bar successfully, cut the top side of the 
bar away down to the rebate line on the end required 


224 


MODERN CARPENTRY 


to be bent. The length of the part cut away will be 
from the bottom of the case to a little beyond the spring¬ 
ing line. Care must be taken to cut the two bars for 



the case in pairs. Steam the bars for several houfs and 
then bend them round the mould (Fig. 29) by securing 
the extreme end first with a cleat, as shown at A. Draw 






































226 


MODERN CARPENTRY 


the bar gradually to the mould, secure it in position by 
the cleat B, and leave it to cool for several hours. It is 
better to leave it on the mould until the following day, 
when the strip to form the rebate—which replaces the 
part cut away—can be fitted and glued in position. 

After the bar has been bent and the strip cleaned 
off, place it on the drawing-board and set out the posi¬ 
tion of the joints at the bottom of the case and on the 
back rail, as already described. 

CIRCULAR-FRONTED CASE WITH SOLID ENDS. 

It will only be necessary, after the preceding explana¬ 
tions, to notice the joint of the back rail, and the section 



1 


n g . 30. 


of the solid end. Fig. 30 shows a section through the 
solid end of the case, grooved to receive the glass. Fig. 31 
is a plan of the angle formed by the end of the case and 
the back rail. The clamp A is tongued and grooved 
to the end, the tongue being stopped ]/ 2 inch below the 
top edge. The clamp is prepared with a hook-joint as 
shown by the dotted lines. The width of the clamp is 
the width of the back rail less the rebate for the glass. 







CASE MAKING 


227 


Fig. 32 shows in isometrical projection the joint at 
the junction of the back rail with the solid end. Imagine 




that A A are brought together. It will then be seen 
that they slide into position and present the appearance 










228 


MODERN CARPENTRY 


shown in the plan in Fig 31, and give the extra lines 
for setting work. 

The solid ends are $4 inch thick, finished size. They 
must be left wide enough to screw to the bottom frame 
of the case. Fix the moulding round the bottom and 
mitre it at each inside round of the ends, as before 
described for upright angle-bars, turning the round on 



the outside of each end out inch above the moulding. 
The moulding mitred round the ends of the case must 
be reduced by the thickness of the quarter-round mem¬ 
ber which forms the rebate for glass at the front of the 
case. 

These cases are often fitted with several trays, the 
bearers to carry them being screwed to the ends. 




SOME FORMS OF PANELS. 


We conclude this Volume by giving some illustrations 
of panels. In Fig. i we give a “flush” panel for a front 
or entrance door, in which in front elevation a, b, are 
the two rails, d d, e e, the stiles, c c, g g, the panel with 



Fig. 1. 



stuck-on mouldings all round and mitring at corners; 
g h is a vertical section in line 3 4. In this the recess 
between the stile and panel is one side only. Where 
there are recesses on both sides of the panel b b, Fig. 
2, and the stiles a a, the panel is known as a “square” 
panel. In this figure the lower diagram is front ele¬ 
vation; that on the left is a section on line 3 4. In 
Fig- 3 we illustrate different forms of panels. In the 
upper diagram, a a, the stiles carry one “square panel,” 
229 ^ 
























230 


MODERN CARPENTRY 




Fig. 3. 































































































































PANELS 


231 


which is not flat, as in Fig. 2, on the inner side, but 
tapers to the centre, which is thickest, to the sides, 



where it may be either square, as at the right hand, 
or finished with a moulding, as on the left. 

Resuming our description of the drawing named, the 
second diagram shows a “flush panel,” with stiles d d, 



the panel having a raised position in the centre, as 
shown at a b, with flat spaces as at c c, all round. The 






































9 


2 $2 MODERN CARPENTRY 

lower diagram to the right is an enlarged view in sec¬ 
tion and elevation of the part of the panel of upper 
diagram to the right. The lower diagram to the left 
is an enlarged view of the left hand side of the panel, 
which is technically called a “raised panel.” Figs. 12 
and 13 are other views of raised panels, and diagram 



Fig. 6. 


B in next figure shows a form of panel in the Gothic. 
Other forms are illustrated in Figs. 8, 9, 10, and 11. 
In Fig. 3 the flat part of the panel surrounding the 
raised central part is called the “margin.” (See also 
Fig. 12 at b.) The panel, as in Fig. 3, is called a 
“moulded raised panel” when there is a moulding at 


































































PANELS 


233 


the margin, as f e h. There are other distinctions in 
panel work, yet to be noticed. In “flush panels/’ as 
in Fig. 1, the “moulding” or “bead” is worked only on 
the two sides (vertical) of the panel, as at d d, Fig. 5, 
and these terminate at the rails, as at f f, no moulding 
being at the ends of the panel. This is called “bead 
butt” panel. When the panel has mouldings all round, 



that is at top and bottom as well as at the sides, the 
mouldings meet at the corners and are mitred, as shown 
in the lower part o’f the diagram in Fig. 6, this is 
known as a “bead flush panel.” In panel work where 
a moulding is worked out of the solid, as at b in Fig. 
4, or at a a in Fig. 5 of the style, as c c or b b, the 
term “stuck on” (a corruption of “struck on,” which 



































234 


MODERN CARPENTRY 


is the true term) is applied. This is only applicable to 
“bead and butt” panel work vertically, as the mould¬ 
ings would not mitre if struck horizontally on the rails. 



When the mouldings are made separately and nailed 
onto the stiles j j, and rails i i, Fig. 6, they are called 
laid on” mouldings. They may be nailed on either to 






































PANELS 


235 








































































































































































































































































236 


MODERN CARPENTRY 




Fig. 12. 





















































































































PANELS 


237 



Fig. 14. 
































































































































































23 8 


MODERN CARPENTRY 


the stiles and rails or to the panels in “flush” work, 
or all around the panels in “square” panels. In Fig. 
14 in diagram A, we give a panel at upper part of 



Fig. 15. 


door, in which the upper rail a a is curved at top, b b b 
the stiles, separated in the centre by a moulding a a d 
the upper panel, with stuck-on mouldings c e e. Dia¬ 
gram B is front elevation of lower panel. In Fig.' 13 


































































































PANELS 


239 


we give a section of middle stile and panel; the middle 
stile b b being provided down the centre with a stuck- 
on moulding, as at b a, corresponding to the vertical 
moulding a a in Fig. 15. A moulding as at c c is worked 
in the margin of the stile corresponding to c c in Fig. 

14. E shows the moulding in section stuck on the square 
panel f g, the margin f being in this way wide. In Fig. 

15, and in Figs. 8, 9, 10, 11 and 12 we give illustrations 
of panel work, and in Fig. 9 section and elevation of 
mouldings for a panel. 


JOINERS’ WORK IN THE CONSTRUCTION OF 
DOORS—DIFFERENT KINDS OF DOORS. 


We now come to illustrate the different forms of 
doors, and various details of their construction. Doors 
are either external or internal and both may be con¬ 



structed much in the same way. The chief difference 
between them, if difference may be made at all, is that 
external doors are heavier in their timbers— that is, 
240 












































CONSTRUCTION OF DOORS 


241 


thicker and broader—and are not quite so much orna¬ 
mented with mouldings, or so highly and carefully fin¬ 
ished, as internal or private room doors. Doors are of 
different classes, beginning with those adapted either for 
houses, of a simple character or for out-buildings, etc., 
where economy is carefully studied, and going up to the 
more elaborate forms, used in houses of the higher class. 

The simplest form of doors is shown in part elevation 
at A, Fig. 16, in plan at B, looking down in direction of 
arrow 1, in C side elevation or edge view looking in 
direction of arrow 2. This form is what is called a 
“batten door.” In elevation in diagram A, the lower 
part is a a, next to floor or ground line b b. The door 
is made up of flat planks, a a c d d, running vertically 
from foot or floor, or ground line b b up to head. These 
are either laid as in plan B in the cheapest class of work, 
edge to edge, and held together by cross pieces, or bars, 
e e. In better work, these and the vertical parts, d d, 
are secured by joints of different kinds. In the section 
C the cross bars e are simply laid flat and nailed to the 
upright planks, d d. The edges of the cross-bars, d d, 
may either be left square, or have the lines or corners 
planed off and “chamfered” or beveled off as at f f. 


BATTEN AND BRACED AND BATTEN, BRACED 
AND FRAMED DOOR. 


Fig. 17 is an elevation in diagram A of a “batten and 
braced” door. To the vertical and cross bars of the 
simple form in Fig. 16 the diagonal -“brace” a a a a, 



corresponding to the struts of a roof truss, are in¬ 
troduced; these butt against the cross bars or battens 
b b b b, while behind are the vertical boards c c c c, 
242 









































































CONSTRUCTION OF DOORS 243 

Diagram B is side elevation or edge view and C vertical 
section. A still higher class of door is the “framed 



Fig. 18. 


braced and battened” door, in Fig. 18, here as in ele¬ 
vation in diagram A, we have an outer frame vertical 






















































M4 


MODERN CARPENTRY 


pieces, held together and secured by the cross-bars 
b, c, d, the ends of these being tenoned into the stiles 
a a. The central spaces are filled with braces e e, and 
the vertical boards f f. Diagram B is vertical section 
on line 2 and C is side view showing ends of tenons of 
cross bars b, c, d; D is plan of top edge, looking down; 
E is cross or horizontal section on line 3 4 in A. 


PANELLED DOORS—NAMES AND OFFICES OF 
DIFFERENT PARTS—STILES—RAILS— 
MORTISES. 

The transition from this form of door to the highest 
class, the “panelled door,” is easy and natural. We have 



seen in the simplest timbers, which is the element of the 
“truss,” and which gives the strongest form attainable. 


2 45 































246 


MODERN CARPENTRY 


In this view the panelled door, as in elevation in A, 
Fig. 19, is not so strong as the form in Fig. 18, from 
the absence of the diagonal braces, as e e, but those, if 
required in a door such as an external one, where 
strength is an object can be dispensed with in interior 
doors, which are always panelled in good houses. 

Elegance or neatness of arrangement, with such or¬ 
namentation as mouldings, etc., can give, are what are 
looked for. In Fig. 19, the external framework enclos¬ 
ing the panels is made up of two side vertical boards, 
a a, b b varying in thickness from 1^ to 2*4 inches, 
and in very superior work even 3 inches. These boards 
are called “stiles”; that by which door is hung to the 
casing, secured by hinges is called the “hanging stile,” 
as a a; that to which the lock is secured the “lock stile,” 
as b b. These stiles are held together by cross-bars 
palled “rails” of which c is the “bottom rail,” d the “top 
rail” and e the “middle or lock rail.” The central ver¬ 
tical bars, as f f are called “muntins” (a corruption of 
mouldings). The assemblage of boards thus arranged 
leaves spaces as g, h, i, j, these are filled with panels, 
as a, b, c and d, in Fig. 20, which is the elevation of a 
/nwr-panelled door. There are also six-panelled doors. 
Generally the panels are nearly equal in length, but in 
some the lower panels are short, the upper being longer. 
Figs. 2 and 4 illustrate outside doors in Continental style. 
The panels are secured to the framing by grooves, as 
shown in preceding figures and as further hereafter 
illustrated, and are ornamented with mouldings, as ex¬ 
plained. In Fig. 19 diagram C is the vertical section, 
edge view of style b b. In Fig. 20 B is plan of top 
edge of door. The rails are secured to the styles by ten¬ 
ons, sometimes single, but more frequently in good work 


CONSTRUCTION OF DOORS 


247 


by double tenons, as in Fig. 21, in which is front elevation 
of rail, a a, b c two tenons. Diagram B is part of stile 
a cut vertically in two to show the seats of the mortises 
b and c, diagram C and view of rail. In left-hand dia- 


e 



3 

Fig. 20. 


gram in Fig. 12 is elevation of part of “lock stile/’ a a 
and “lock rail,” b of a bedroom door, with simple lock, 
c, known as a “rim lock.”' In diagram B, part of the 
“hanging stile,” a a, of this door is given in elevation, b 






























248 


MODERN CARPENTRY 


part of “top rail,” a portion of upper “hinge” is shown 
at c. Diagram C is edge view. The inner edges of 
stiles, rails, and mortises are generally, in good work, 
“stop chamfered” as at d d, or beveled off from end to 
end, as at e f, the two edges meeting in a mitre, as 
shown. The “top chamfer,” d d, is the neatest, stopping, 
as it does, short of the end. A rim lock is screwed 
onto the outside of the lock stile; what is called a 
“mortise lock” is employed in superior doors, where the 
lock is concealed, nothing but the handle and keyhole 
being visible, the lock being inserted in a mortise or 
vacant part cut out in the stile to receive it. Fig. 29 
contrasts the two locks, c d is the rim lock. In the 
mortise lock nothing but the handle at g is seen, and the 
escutcheon h, i is the bolt of the lock, a a, b b, a' a', 
b' b', are the chamfered stiles and rails. 


DOOR CASINGS. 


Doors are secured to “casings.” These are of tim¬ 
ber, and built into the walls, and are secured to wood, 
bricks or grounds. Fig. 23 illustrates in part elevation 
an outer “door casing.” The sides b b, c c, are called 
“jambs,” f f, the “head,” into which the jambs are ten¬ 
oned, the feet being also tenoned, at d, into the upper 
part of stone step a a. Fig. 22 is sectional plan show¬ 
ing arrangement and relative positions of various parts 
of a door and its casings. The door, 1 1 , is hinged to 
the “jamb” b, this being secured to the “ground” or 
“wood brick” a a, bulit into the wall b b, c and j are 
the “architraves.” The opposite “jamb,” f f, is rebated 
as at m to allow of a space into which the “door lock 
stile” falls, as shown by the dotted lines, which repre¬ 
sent the lines of the door. The outer edge of the jamb 
may be left plain, but is often finished ofif with a “quirked 
head,” as at j; k, k, the hinge. The inner and outer 
architraves are at c and j; a a, the wood brick; b b, the 
wall; e, i, are the elevations of the architraves, d and h. 
The elevations of these two parts of sectional plan of 
door fittings are given in the under part of the drawing 
in Fig. 23. The edge of the door a, as looking at it 
from the inner side, is shown at p p, q q, being the 
ends of tenons of top rail, r r, the hinge, n n, from a view 
of architrave, o o the wall in the void of which the door 
is hung. In the under drawing to the right, part of 
front surface of door is shown, s s, the architrave, t t 
the wall. 


249 


250 


MODERN CARPENTRY 



Fig- 22 - Fig. 23. 
















































































































JOINTS OF STILES AND RAILS IN PANELLED 
DOORS. 

Figs. 24 and 25 give illustrations of methods of join¬ 
ing rails and stiles, or rails and mortises. Let abed, 
Fig. 24, be the stile, with moulding stuck on edge; f g h 
is part of the rail, with tenon f, shown by dotted lines 



Fig. 24. 


in stile abed. Front view of tenon are face of mitre 
of chamfer at p, looking at a b c d in the direction of 
arrow 1, is shown in the lower diagram at k', p' and e". 
The section of part f g looking at its end, in direction of 
arrow 2, is shown at 1 m n; the section of a moulding 

251 







































252 


MODERN CARPENTRY 


is in this at e'. In lower diagram to the right is given 
a view of under side of rail f g. In Fig. 25, a a, is 



Fig. 25. 


front view of part of stile with moulding worked on 
edge, at b b; part of rail is at c' c' d. The angular 
































































STILES AND RAILS 


253 


face of part cut out in stile e f, fg corresponds with 
angular end h i j of rail, but a tenon i 1 k is left on, or 
is inserted in end of piece c' c' d. The end view of the 
stile a a, looking at it in the direction opposite to that 



Fig. 26. 


of the arrow 3, is shown in the middle diagram to the 
right with corresponding letters accented, showing cor¬ 
responding parts. The line i" i" corresponds to the line 
at point in rail c' c' d d. The plan of under side of rail 
c' c' d is shown in diagram immediately below k', 1 be¬ 
ing edge view of tenon k 1 . The finished joint is shown 
at o o, p p; the diagram below to the left being cross 



Fig. 27. 


section to the line 1 2. Enlarged elevation q, and sec¬ 
tion r of moulding b b, or b", is given at the two dia¬ 
grams to the right at bottom of drawing. Another 
method of forming the junction is shown in the middle 
diagram at the foot of Fig. 25, the shaded part showing 
form of tenon with the ends of moulding united. 













A FOUR-PANELLED DOOR. 


In Fg. 28 I give a drawing—to a scale of y$, or iy 2 
inch to the foot—of a four-panelled interior or room 



door, showing all the leading parts of the framework, 
with the exception of top rail, which is usually about 
half the breadth or depth of the middle of lock rail, 
marked b b in the drawing. The panels are not shown, 
254 
















































STILES AND RAILS 


255 


but the dimensions of the spaces they occupy are given. 
The panels are plan “square,” the only ornamentation in 
this example being a “stop chamfer” worked on the mar¬ 
gin of stiles, and rails, as shown at g g and h h. In the 
drawing a a is the “bottom rail,” b b the middle, or usu¬ 
ally “lock rail,” as it carries the “mortise lock,” the 
handle of which is shown at j. The “key hole” is 
covered by a movable part, hung or jointed at upper 
end, called the “escutcheon,” or more frequently in tech¬ 
nical talk, the “scutcheon,” or “skutcheon,” shown at k. 
The stiles are at c c, e e—the stiles c c, termed the “lock 
stile,” being that in which the lock is mortised. The 
stile e e is called the “hanging stile,” being that on 
which the door is “hinged” or “hung” to the door casing. 
The vertical pieces, or “muntins,” which divide the pan¬ 
els from each other, placing them in pairs on each side 
of the door, are shown at d d. The door framing thus 
constructed is surrounded on both sides and at top by 
the architraves f f f. 


ARCHITRAVES OF A FOUR-PANELLED DOOR. 


The section of architrave in relation to the door cas¬ 
ing or check is in upper diagram to the left in Fig. 29, 
a a being part of the door casing, b b the section of 
architrave, of which part elevation is shown at c c, I, 
2, 3, and 4 showing similar parts in section correspond¬ 
ingly lettered. The edge view of the “lock stile” as a a f in 
the figure preceding, is shown at d d; e e shows the brass 
plate let into the edge and secured by screw nails as 
shown. This is part of the lock furniture of the door, 
f indicating position and section of the shooting or lock¬ 
ing bolt of the lock, which passes into tEe aperture of a 
brass plate secured to the inner side or edge of the door 
casing. The bolt, which secures the door, being closed 
—not locked—f being the locking bolt, is shown at g, 
this being worked by the handle j of the lock. The part 
of the lock furniture attached to the door casing oppo¬ 
site to the edge, as d d d, of the door stile, is shown in 
the lower diagram to the right. The part 3 3 in this 
corresponds to the face of the recessed or rebated part p 
in drawing above, cut in the face of the door casing n n n, 
the door passing into and resting against the face of re¬ 
cess or rebate p. In the upper diagram to the right, 
o o o is the outer architrave secured to the door casing 
n n n, r part of the inner architrave. The part of the 
lock furniture secured to the door casing is shown at 
t t; it is a brass plate let into the face g, or 3 3 of recess 
or rebate p. The aperture in this into which the bolt f 
of the lock passes is shown at p; that into which the bolt 


ARCHITRAVE OF DOOR 


257 



Fig. 29. 






























































258 


MODERN CARPENTRY 


moved by the hand passes is at u, a spring w, cast onto 
the plate t t, being shown at w. A small projecting part 



Fig. 30. 


as w', to make the opening and closing of the door more 
easy. The two diagrams to the left at lower part of 



drawing show the elevation k 1 m, the chamfered part of 
framing with section at k' k\ 





























































SOME EXAMPLES OF ORNAMENTAL WOOD¬ 
WORK. 


The following examples are introduced in order to 



give the workman an idea of the shape and construc¬ 
tion of low-cost ornamental wood-work. The figures 
259 






















260 


MODERN CARPENTRY 



PANUINC 



AAiSEO PANU 




shown from No. i to No. 12, inclusive, exhibit a num¬ 
ber of large boards, chiefly in Gothic style. Plate No. 2 is 
a style which was in vogue very much a few years ago 



























































































































ARCHITRAVE OF DOOR 261 

and was generally known among carpenters as Ginger- 
Bread work. It is well adapted for sea-side cottages or 



No. 13. 

summer residences; it consists mostly of cutwork. Nos. 
2, 5, 8 and 9 are well adapted for ordinary cottage 



No. 14. 

work. Nos. 13, 16 and 21 are well suited for balus¬ 
trades, No. 16 being especially adapted for heavy balus- 



No. 15. 

trades on verandas or over bay windows. Nos. 14, 15 
and 17 require no explanation, as they may be adapted 






















MODERN CARPENTRY 






















































ARCHITRAVE OF DOOR 


263 

to a thousand different purposes. Nos. 18 and 19 make 
very handsome drops for verandas and other similar 



No. 17. 

work. No. 29 shows a single drop with the grain of 
the wood running vertically. A number of these placed 



No. 18. 

together edge to edge make a very nice trimming for 
verandas. No. 22 shows a cut bracket which will often 



No. 19. 

be found useful. No. 23 shows an elaborate railing 
suitable for a veranda or balcony. No. 24 exhibits a 











264 


MODERN CARPENTRY] 



No 20. No. 21. 



No. 22. 





















































































































ARCHITRAVE OF DOOR 


265 



No. 23. 




No. 25. 


























266 


MODERN CARPENTRY 


perforated panel suitable for many places. No. 26 
shows a portion of a circular panel which may be per¬ 
forated or the ornaments may be planted on, according 
to exigencies. See Plates. The balance of the examples 
shown speak for themselves. They offer a number of ex¬ 
cellent suggestions to the progressive workman. These 
examples will doubtless prove of great value to the work¬ 
man. 


QUESTIONS ON MODERN CARPENTRY 
VOL. I. 


QUESTIONS. 

1. Give definition of a “circle.” 

2. What term is given to a line that is drawn through 
center to circumference of a circle? 

3. What term is given to a line drawn from center 
to circumference of a circle? 

4. What term is given to a line (less than the diam¬ 
eter) that cuts the circumference of a circle at two 
points ? 

5. Give definition of a “tangent.” 

6. Give definition of a “segment of a circle.” 

7. Give sketch of a circle showing the “diameter,” 
“radius,” “chord,” “segrnent” and “tangent.” 

267 


268 


MODERN CARPENTRY 


8. Give sketch and describe how to find the center 
of a circle. 

9. Into how many equal parts is the measurement 
of the circumference of a circle divided ? 

10. Give the three terms used in measurement of 
the circumference of a circle, and show how they are 
written. 

11. What is a quadrant of a circle? 

12. How many degrees are in a quadrant of a circle? 

13. How many degrees are in a semi-circle? 

14. What term is given to the angle of a circle that 
is half of a right angle? 

15. Give sketch and describe how three right angles 
may be formed within a semi-circle. 

16. Give sketch and describe how a hexagon may 
be formed within a circle. 

17. Give sketch of a hexagon showing how an equi¬ 
lateral triangle may be formed. 

18. Give sketch and describe how a right angle or 
quadrant may be bisected. 

19. Give sketch and describe how to get a straight 
line that shall equal the circumference of a circle or part 
of a circle or quadrant. 

20. Give sketch and show how quadrant may be di¬ 
vided into any number of equal parts, say thirteen. 

21. Give sketch and show how equilateral triangle 
may be employed in forming the trefoil. 

22. Give sketch and describe method of finding the 
“stretch out” or length of circumference of a circle. 

23. Give rule by arithmetic of how to find the cir¬ 
cumference of a circle. 

24. Give sketch and describe how a curve having 


QUESTIONS 269 

any reasonable radius, may be obtained, if but three 
points in the circumference are available. 

25. Give a practical illustration of how to find a place 
to locate a center, where the diameter is great. 

26. What is a “polygon?” 

27. Give the names applied to polygons having three 
sides, four sides, five sides, six sides, seven sides, eight 
sides, nine sides, ten side, eleven sides, and twelve 
sides respectively. 

28. Give the two names under which polygons are 
classified. 

29. Give sketch showing how a trigon may be con¬ 
structed and how the miter joint may be obtained. 

30. Give sketch and describe how a square may be 
formed. 

31. Give sketch and describe how to construct a 
pentagon. 

32. Give sketch and describe how a hexagon may 
be formed. 

33. Give sketch and describe how a heptagon may 
be formed. 

34. Give sketch and describe how an octagon may 
be formed. 

35. Show practically how all regular octagons may 
be constructed. 

36. Give a practical illustration of how a perpendicu¬ 
lar line may be made on any given straight line. 

37. Give a practical illustration of how to bisect an 
angle by the aid of the steel square alone. 

38. Give a practical illustration of how to bisect an 
acute angle by same method—steel square. 

39. Show practically how to get a correct miter cut, 
or angle of 45 0 on a board. 


270 


MODERN CARPENTRY 


40. Show how to construct a figure showing an 
angle of 30° on one side, and on the other an angle 
of 6o°. 

41. Show how the diameter of a circle may be ob¬ 
tained through the aid of the steel square. 

42. Show how an equilateral triangle may be ob¬ 
tained through use of the steel square. 

43. Show how to describe an octagon by using the 
steel square. 

44. Show how a near approximation of the circum¬ 
ference of a circle may be obtained by use of the steel 
square and a straight line. 

45. Give illustration how a board may be divided into 
any given number of equal parts by aid of steel square 
or pocket rule. 

46. Give the definition of an “ellipse.” 

47. Give an illustration of one of the simplest meth¬ 
ods of describing an ellipse. 

48. Give an illustration of projecting an ellipse by 
using a trammel. 

49 - Give illustration of describing an ellipse by the 
intersection of lines. 

50. Give illustration of describing an ellipse by the 
intersection of arcs. 

51. Show how radial lines may be obtained for arches 
and elliptical work. 

52. Give an illustration how to describe a diamond 
or lozenge-shaped figure. 

53. Give illustration how to describe a spiral or 
scroll by a simple method. 

54. Give illustration of how a spiral may be described 
in a scientific manner, and which can be formed to di¬ 
mension. 


QUESTIONS 


271 


55. Give illustration of the method of obtaining a 
spiral by arcs of circles. 

56. Give illustration and method of forming a “par¬ 
abola.” 

57. Give illustration and method of forming an “hy¬ 
perbola.” 

58. Give the names of the different kinds of arches 
in buildings. 

59. Mention the names given to pointed arches. 

60. What is the name given to the stones forming 
an arch? 

61. What is the name given to the centre stone in 
an arch? 

62. Give the names applied to the various divisions 
of an arch, namely, the highest point, the lowest point, 
and the spaces between respectively. 

63. What is the name given to the under or concave 
surface of an arch ? 

64. What is the name given to the upper or convex 
surface of an arch? 

65. What are the names given to the supports of an 
arch? 

66. Show by illustration and describe how to obtain 
the curves and radiating lines of a semi-circular arch. 

67. Show by illustration and describe how to obtain 
the curves and radiating lines of a segment arch. 

68. Show by illustration and describe two examples 
of Moorish or Saracenic arches, one of which is pointed. 

69. What is a “flatband”? 

70. Give illustration and describe how to obtain 
the curves and radiating'.lines of the elliptic arch. 

71. Give illustration and describe how the centers 
and curves of an equilateral arch may be obtained. 


272 


MODERN CARPENTRY 


72. Give illustration and describe how the centers 
and curves of a lancet arch may be obtained. 

73. Give illustration and describe how the center and 
curves of a low or drop arch may be obtained. 

74. Give illustration and describe how the centers and 
curves of a Gothic arch with a still less height, may be 
obtained. 

75. Give illustration and describe another four-cen¬ 
tered arch of less height. 

76. Give illustration and describe how to obtain an 
equilateral Ogee arch. 

77. Give illustration and describe method of obtain¬ 
ing the lines for an Ogee arch, having a height equal to 
half the span. 

78. Give some instances in carpenter work where 
half of the Ogee curve is employed. 

79. Give a description of the steel square and its 
several divisions. 

80. Give a practical illustration of how a board or 
scantling may be measured by use of steel square. 

81. Give rule how to find hypothenuse of a right- 
angled triangle. 

82. Give an illustration of how length of braces may 
be obtained by use of the square. 

83. Describe the use of the “octagonal scale” on the 
tongue of the square. 

84. Show method how the pitch of a roof may be ob¬ 
tained by use of the square. 

85. Show method to obtain bevels and lengths of hip 
rafters by use of the square. 

86. Show method for finding the length and cuts for 
cross-bridging. 


QUESTIONS 273 

87. Show method for obtaining the “cuts” for octa¬ 
gon and hexagon joints. 

88. Show by illustration the method of defining the 
pitches of roofs, and giving the figures on the square 
for laying out the rafters for such pitches. 

89. Give a short description of what is known as bal¬ 
loon framing, and how the different parts are con¬ 
structed. 

90. Give illustration and describe a “hip-roof.” 

91. Give illustration and describe a “lean-to-roof.” 

92. Give illustration and describe a “saddle-roof.” 

93. Give illustration and describe a “mansard roof.” 

94. Give illustration and describe a simple hip-roof 
having a ridge. 

95. Give illustration and describe an “octagon roof.” 

96. Give illustration and describe manner of con¬ 
struction of a “dome roof.” 

97. Give illustration and rules for construction of an 
octagonal spire. 

98. Give a few illustrations of scarfing timbers. 

99. Show a few examples of strengthening and trus¬ 
sing joints, girders and timbers. 

100. Explain what is meant by the term “kerfing.” 

101. Give illustration showing how to determine the 
number and distances apart of saw kerfs required to 
bend a board round a corner. 

102. Give illustration of how to make a “kerf” for 
bending round an ellipse. 

103. Describe how to bend thick stuff around work 
that is on a rake. 

104. Give illustration 'and describe how to lay out 
a hip rafter for a veranda having a curved roof. 

105. Give illustration and describe how to obtain 


274 


MODERN CARPENTRY 


the curve of a hip rafter, when the common rafters have 
an ogee or concave and convex shape. 

106. Give illustration and describe how raking mould¬ 
ings are used to work in level mouldings. 

107. Describe the kind of mouldings called “spring 
mouldings.” 

108. Give illustrations showing plan and elevation of 
cluster column of wood for 4 columns and describe how 
constructed. 

109. Give illustration of a hopper and describe how 
to be constructed. 

no. Give illustration and describe how a conical 

tower roof may be curved. 

in. Give illustration and describe how to cover a 
dome roof. 

112. Give illustration and describe how the semi¬ 

circular soffit of a doorway may be made. 

113. Describe how a circle soffit may be laid off into 
panels. 

114. Give illustration and describe method for ob¬ 
taining correct shape of a veneer for a gothic-splayed 

window or door head. 

115. Give illustrations and describe two methods of 
dovetailing hoppers, trays and other splayed work. 

116. Give description of how an ordinary straight 
flight of stairs may be constructed. 

117. Give sketch showing part of a straight stair. 

118. Give sketch showing stair with winders and 
landing. 

119. Give sketch and describe a stair with brackets. 

120. Give sketch showing stair with two newels and 
balusters, also paneled string and spandrel. 


QUESTIONS 


275 


121. Give sketches of seven of the latest designs for 
doors. 

122. Give five sketches showing methods of con¬ 
structing and finishing a window frame for weighted 
sash. 

123. Give sketches showing the various parts of a 
bay window for a balloon frame. 

124. Give illustration and describe six examples of 
shingling roofs. 

125. Show by sketch how panels are formed. 

126. Describe the various kinds of panels named. 

127. Make sketch of a four-panel door. 

128. How are air-tight cases made? Describe the 
method of making. 

129. What is meant by the word “stile” ? 

130. What is a rail in a door ? What is a muntin ? 0 

131. What is a chamfer? Describe one. 

132. Examine examples of sketches of ornamental 
wood-work, draw and describe a “baye-board.” 

133. Make a design of perforated insular panel. 


INDEX 


A 

Airtight wall case.195 

Angle bars at different angles.131 

Arches, elliptical . 63 

Flat. 62 

Four centered. 65 

Horseshoe. 62 

Lancet . 64 

Lintel . 63 

Ogee . 66 

Segmental . 61 

Architrave of door.263 

B 

Balloon framing, description of.173 

Balusters and turned work. 68 

Bay window frames, sections of.174 

Bending blocks for splayed heads.153 

Bevels and cuts for rafters. 83 

Bevels for hips, jack rafters and purlins. 85 

Bisecting angles . 28 

Bisecting angles with steel square. 34 

Boxes for different measures, sizes of.185 

Brace rule..73 

Braces, table of. 75 

Brick work . 187 


























INDEX 


c 

Case making. 195 

Centers, finding . 20 

Circle, the . 9 

Circular door entrances.152 

Cisterns, capacity of.192 

Cluster columns.133 

Cluster columns, bases and capitals of.133 

Cornices on a rake, inside.132 

Cutting bridging . 78 

Cutting raking mouldings in miter box.130 

Cycloidal curves ..... 57 

D 

Degrees . 14 

Dividing lines. 79 

Door, architrave of.263 

Architrave of a four paneled.256 

Casings.249 

Doors, batten and braced.242 

Construction of .240 

Description of.170 

Four paneled.254 

Joints of stiles and rails.251 

Names of different parts.245 

Paneled .245 

Styles of .169 

Dovetailing .156 

Blind .159 

Common.157 

Lapped. 158 

Splayed .159 































INDEX 


E 

Ellipses, spirals and other curves 
Elliptical curves, description of. 
Excavations. 


Flashings for valleys. 

Flexible radial guide. 

Framing corners, etc. 

Sills, etc... 

Triangular. 

H 

Hopper cuts, housed. 

Hopper lines, compound. 

Hoppers, butt cuts for. 

Corner blocks for. 

Corner blocks for obtuse. 

Miter cuts for. 

Miters for obtuse. 

Miters for square. 

Regular . 

J 

Jack rafters, lengths of. 

Joiner’s work generally. 

K 

Kerfed stuff, bending....... 

Kerfing for an ellipse....... 

Kerfing on a rake. 

Kerfs, laying out. 


. 41 

46 

.184 

177 

49 

91 

87 

109 

137 

143 

136 

139 

140 

135 

142 

141 

134 

102 

167 

119 

120 

121 

118 

























INDEX 


L 

Laying out curved hips.123 

Curved hips and jack rafters.125 

Ogee hips and rafters.124 

Raking mouldings for circular pediments.129 

Loads, safe-bearing .192 

Lumber, measurement table.182 

Rule . 71 

M 

Masonry . 135 

Materials for roofs, weight of.189 

Strength of .182 

Miscellaneous illustrations .172 

Mitering circular and straight mouldings.122 

Circular mouldings .121 

Curved mouldings in panels.122 

Mortise and tennon in timber.Ill 

Mouldings . 67 

N 

Nails and tacks, number per pound.193 

Number of required in carpentry work.185 

0 

Octagon rule on steel square. 76 

Octagons . 30 

Ornamental woodwork ... u.259 

Ornamentation . lg 

Ovals . 50 


























INDEX 


P 

Panels, forms of.229 

Parabola and its uses. 56 

Pitch-board and strings.160 

Pitches, laying off. 81 

Polygons . 22 

Q 

Questions for students.267 

R 

Rafter rule by steel square. 76 

Raking mouldings . .....126 

Raking mouldings for pediments.128 

Reinforcing timber.113 

Roof, core for conical.150 

Covering of a conical.149 

Inclined domical.151 

Roofs and roofing generally. 96 

Covering domical .150 

Domical... 105 

Lines for hip.98 

Octagon hip . 99 

Sisser.104 

Trussed .103 

S 

Setting rail and newel post. .164 

Shingles, table for estimating. .184 

Shingling.184 

Different methods.175 

Hip rafters.175 



























INDEX 


Shingling—Continued. 

Illustrations of. 176 

Yalleys . 175 

Siding, flooring and laths.184 

Slates, number of, required per square yard.188 

Slating. 188 

Snow and wind loads.189 

Soffits, Gothic .;. 155 

Splayed .:.154 

Solutions of problems with steel square. 34 

Spirals . 52 

Spires and spire framing.108 

Steps bracketed .165 

Method of forming .165 

Stair building.159 

Stairs, dog-legged.162 

Open string.164 

Various styles of.168 

Winding.163 

Steel square, description of. 70 

Straight line solutions . 32 

Strapping timber . 112 

Superficial or flat measure, table of.183 

T 

Tangents. 11 

Timber scarfing .110 

Timber measure, round and equal-sided.183 

Treads and risers, table of.163 

Treads, risers and strings.161 

Trimming stairs, chimneys) etc. 89 

Trussing and strengthening timber.114 

Turned mouldings and carved newels.174 
































INDEX 


W 

Weights and measures, cubic or solid.191 

Land measure.199 

Linear measure .191 

Miscellaneous measures.191 

Square measure .191 

United States measure.... • 190 

Window frames and sections.171 

Wind pressure on roofs.193 










MODERN CARPENTRY 

VOLUME II. 





PART 1. 

SOLID GEOMETRY. 

INTRODUCTORY. 

In the first volume of Modern Carpentry I gave a 
short treatise on plain or Carpenters’ Geometry, 
which I trust the student has mastered, and thus pre¬ 
pared himself for a higher grade in the same science, 
namely, Solid Geometry: and to this end, the follow¬ 
ing treatise has been selected, as being the most simple 
and the most thorough available. 

Solid geometry is that branch of geometry which 
treats of solids—i.e., objects of three dimensions 
(length, breadth and thickness). By means of solid 
geometry these objects can be represented on a plane 
surface, such as a sheet of paper, in such a manner 
that the dimensions of the object can be accurately 
measured from the drawing by means of a rule or 
scale. The “geometrical” drawings supplied by the 
architect or engineer for the builder’s use are, with 
few exceptions, problems in solid geometry, and there¬ 
fore a certain amount of knowledge of the subject is 
indispensable, not only to the draughtsman who pre¬ 
pares the drawings, but also to the builder or work¬ 
man who has to interpret them. 

As the geometrical representations of objects con¬ 
sist entirely of lines and points, it follows that if pro¬ 
jections of lines and points can be accurately drawn, 
9 


10 


MODERN CARPENTRY 


the representation of objects will present no further 
difficulty. A study of lines and points, however, is 
somewhat confusing, unless the theory of projection 
has first been grasped, and for this reason the sub¬ 
ject will be introduced by a simple concrete example, 



•Vertical Projections (or Elevations) and Horizontal Projection (or Plan) of a 
Table in the Middle of an Oblong Room 

Fig. 1. 


such, as a table standing in the middle.of a room. The 
four legs rest on the floor at A, B, C, and D (Fig. 1), 
and the perpendiculars let fall from the corners of 
the table-top to meet the floor at E, F, G, and H. The 
oblong E F G H represents the horizontal projection 
or “plan” of the table-top; the small squares at A, 





























SOLID GEOMETRY 


11 


B, C, and D represent the plan of the four legs, and 
the lines correcting them represent the plan of the 
frame work under the top. The large oblong J K L M 
is a plan of the room. 

The plan or horizontal projection of an object is 
therefore a representation of its horizontal dimensions 
—in other words, it is the appearance which an object 
presents when every point in it is viewed from a posi¬ 
tion vertically above that point. 

In a similar manner an elevation or vertical projec¬ 
tion of an object is a representation of its vertical di¬ 
mensions, and also, it should be added, of some of its 
horizontal dimensions,—i. e., it is the appearance which 
an object presents when every point in it is viewed 
from a position exactly level with that point, all the 
lines of sight being parallel both horizontally and ver¬ 
tically. Thus, the front elevation of the table (or the 
vertical projection of the side G H) will be as shown at 
G' H' C' D', and the vertical projection of the side J K 
of the room will be JKJK. The vertical projection 
of the end of- E G of the table will be as shown E" 
G" A" C" and of the end K M of the room will be 
K" M" K M; the drawing must be turned until the line 
K M is horizontal, for these and projections to be 
properly seen. 

By applying the scale to the plan, we find that the 
length of the table-top is six and a half feet, the 
breadth 4 feet, and the distance of the table from each 
wall 4% feet. From the front elevation we can learn 
the height of the table, and also give its length and dis¬ 
tance from the end walls of the room. From the end 
elevation we can ascertain the breadth of the table and 


12 


MODERN CARPENTRY 


its distance from the sides of the room, as well as its 
height. 

To make the drawing clearer, let us imagine that 
the walls of the room are of wood and hinged at the 
level of the floor. On the wall J K draw the front ele¬ 
vation of the table and then turn the wall back on its 
hinges until it is horizontal,—i. e., in the same plane 
as the floor. Proceed in a similar manner with the end 
K M, and we get the three projections of the room and 
table on one plane, as shown in the diagram. To avoid 
confusion the end elevation will not be further con¬ 
sidered at the present. 

It will be seen that the line J K represents the angle 
formed by the wall and floor,—in other words, it repre¬ 
sents the intersection of the vertical and horizontal 
“planes of projection,” it is known as “the line of in¬ 
tersection,” or “the ground line.” If a line is let fall 
from G' perpendicular to J K, the two lines will meet 
at G, and they will be in the same straight line. Simil¬ 
arly, the perpendiculars H' li and h H are in the same 
straight line. Lines of this kind perpendicular to the 
planes of projections are known as “projectors” and 
are either horizontal or vertical G' g and H' h are ver¬ 
tical projectors; G g, C c, D d, and Hh are horizontal 
projectors. 

Vertical projectors are not always parallel to one of 
the sides of the object represented, or, if parallel to 
one side, are not parallel to other sides which must 
be represented; thus, a vertical projector or “eleva¬ 
tion” of an octagonal object, if parallel to one of the 
sides of the octagon, must be oblique to the two adja¬ 
cent sides. In Fig. 2 an octagonal table is shown. The 


SOLID GEOMETRY 


13 


plan must first be drawn, and from the principal 
points of the plan projectors must be drawn perpen¬ 
dicular to the vertical plane of the projection, until 
they cut the ground line, and from this perpendiculars 
must be erected to the height of the several parts of 

A B CD 



tPlan and Elevation of an Octagonal Table 

Fig. 2. 

the table. The elevation can then be completed with¬ 
out difficulty. The side B C of the table is parallel 
to the vertical plane of projection, but the adjacent 
sides AB and CD are oblique.* 

*A distinction must be made between “perpendicular” and 
“vertical.” The former means, in geometry, a line or plane at 
right angles to another line or plane, whether these are hori¬ 
zontal, vertical or inclined; whereas a vertical line or plane 
is always at right angles to' a horizontal line or plane. The 
spirit-level gives the horizontal line or plane, the plumb-rule 
gives the vertical. 




























14 


MODERN CARPENTRY 


2. POINTS, LINES, AND PLANES. 

I. To determine the position and length of a given 
straight line, parallel to one of the planes of the pro^- 
jection. 

Let GH (Fig. 3) be the given straight line. To de¬ 
termine its position (i e., in regard to horizontal and 
vertical planes), it is only necessary to determine the 
position of any two of the extreme points. 


A 

CD 

C 

o 


8 ' 

E 

F 


o 3 & 9 in. 

i ■ ■ i ■ ■ i ■ . t ■ . i 



•No. 1. Perspective View. No. 2. Geometrical Projections 
Horizontal and Vertical 

Fig. 3. 

Let A B C D be a vertical plane parallel to the given 
line, and C D E F a horizontal plane. The vertical pro¬ 
jection or elevation of the line is represented by the 
line h g, and the horizontal projection or plan by the 
point g', the various projections being shown by dotted 
lines. The given line is proved to be vertical, because 


















SOLID GEOMETRY 


15 


its horizontal projection is a point; its length as meas¬ 
ured by the scale, is 6 inches; its height (gg") above 
the line of intersection C D is 4 inches; and its hori¬ 
zontal distance (g' g") from the same line is 8 inches. 

If the illustrations are turned so that C D E F be¬ 
come a vertical plane, and A B C D the horizontal plane 
then G H will be horizontal line, because one of its 
vertical projections is a point. Other vertical projec¬ 
tions of the line can be made,—as, for example, a side 
elevation,—in which the projection will appear as a 
line and not a point, but a line must be horizontal if 
any vertical projection of it is a point. 



Fig. 4. 

Let the given line GH (Fig. 4) be parallel to the 
vertical plape, but inclined to the horizontal plane. 
Then gh will be a vertical projection, and g'h' its 
horizontal projection or plan. By producing h g till it 
cuts CD at c, it will be found that the given line is 
inclined at an angle of 60° to the horizontal plane; 
its length, as measured by the scale along the vertical 































16 


MODERN CARPENTRY 


projection gh, is 8 inches; the height of Gr above the 
horizontal plane (measured at gg") is 3 inches, and 
the height of H (measured at hli") is 9% inches. 

II. To determine the length of a given straight line, 
which is oblique to both planes of projection. 



Let ab (Fig. 5) be the horizontal projection and 
a' b' the given vertical projection of the given line. 
Draw the projection A B on a plane parallel to ab in 
the manner shown. From this projection the length 
of the given line will be found (by applying the scale) 
to be 10 inches. The height C B is of course equal to 
the height c' b', and the horizontal measurement A C 
is equal to the horizontal projection a b, and the angle 










SOLID GEOMETRY 


17 


A C B is a right angle. It follows therefore that, if from 
b the line b d is drawn perpendicular to a b and equal to 
the height c' b', the line joining a d will be the length 
of the given line. To avoid drawing the horizontal line 
a' c' the height of a' and b' above xy are usually set 
up from a and b as at a" and b", the line a" b" is the 
length of the given line.* 

III. The projections of a right line being given, to 
find the points wherein the prolongation of that line 
would meet the planes of projection. 



-I. Perspective View; II. Vertical and Horizontal Projections 


Fig. 6. 


Let ab and a'b' (Fig. 6, II) be the given projection 
of the line AB. In the perspective representation of 
the problem it is seen that A B, if prolonged, cuts the 
horizontal plane in c, and the vertical plane in d, and 
the projections of the prolongation become ce and 
f d. Hence, if ab, a'b' (Fig. 6, II) be the projections 

♦The application of this problem to hips is obvious. Sup¬ 
pose that a b is the plan of a hip-rafter, and a' b' an eleva¬ 
tion, the length of the rafter will be equal to a d or a b. 












18 


MODERN CARPENTRY 


of A B, the solution of the problem is obtained by pro¬ 
ducing these lines to meet the common intersection of 
the planes in f and e, and on these points raising the 
perpendiculars f c and e d, meeting a b produced in c 
and a' b' produced in d; c and d are the points 
sought. # 

IV. If two lines intersect each other in space, to 
find from their given projections the angles which 
they make with each other. 

Let a b, c d, and a' b' c'd' (Fig.7) be the projections 
of the lines. Draw the projectors e' f' f' e', perpen¬ 
dicular to the line of intersection a' c', and produce 
it indefinitely towards E"; from e draw indefinitely, 
e E' perpendicular to the line e' f, and make e E' 
equal to f" e, and draw fE'. From f as a centre 
describe the arc E" gE" , meeting e f produced in E", 
and join a E" c E". The angle a E" c is the angle 
sought. This problem is little more than a develop¬ 
ment of problem II. If we consider e f, e'f', as the 
horizontal and vertical projections of an imaginary 
line lying in the same plane as a e and c e, we find 
the length of this line by problem II to be f E'; in 
other words f E'is the true altitude of the triangle 
a e c, a' e' c'. Construct a triangle on the base a e 
with an altitude f E' equal to f E', and the problem is 
solved. The practical application of this problem will 

*The points d and c are known respectively as the vertical 
and horizontal “traces” of the line a b, the trace, of a straight 
line on a plane being the point in which the straight line, 
produced if necessary, meets or intersects the plane. A hori¬ 
zontal line cannot therefore have a horizontal trace, as it can¬ 
not possibly, even if produced, meet or intersect the horizontal 
plane; for a similar reason, a vertical line cannot have a ver¬ 
tical trace. 



SOLID GEOMETRY 


19 


be understood if we imagine a e c to be the plan and 
a' e' c' the elevation of a hipped roof; f E' gives us the 
length and slope of the longest common rafter or spar, 
and E" c is a true representation of the whole hip, i. 
e., on a plane parallel to the slope of the top. It will 



be observed that the two projections (e and e') of the 
point of intersection of the two lines are in a right 
line perpendicular to the line of intersection of the 
planes of projection. Hence this corollary.— The pro- 





20 


MODERN CARPENTRY 


jections of the point of intersections of two lines which 
cut each other in space are in the same right line per¬ 
pendicular to the common intersection of the planes 
of projection. This is further illustrated by the next 
problem. 




•I. Vertical and Horizontal Projections. II. Perspective View 
Fig. 8. 

V. To determine from the projection of two lines 
which intersect each other in the projections, whether 
the lines cut each other in space or not. 

Let a b, c d, a b', c d' (Fig. 8, I) be the projections 
of the lines. It might be supposed, as their projections 








SOLID GEOMETRY 


21 


intersect each other that the lines themselves intersect 
each other in space, but on applying the corollary of 
the preceding problem, it is found that the intersec¬ 
tions are not in the same perpendicular to the line of 
intersection a c of the planes of projection. This is 
represented in perspective in Fig. 8, II. We there 
see that the original lines a B c D do not cut each 
other, although their projections a b, c d, a b', c d', 
do so. From the point of intersection e raise a perpen¬ 
dicular to the horizontal plane, and it will cut the 
original line c D, in E, and this point therefore belongs 
to the line c D, but c belongs equally to- a B. As the 
perpendicular raised on e passes through E on the line 
c D, and through E on the line a B, these points E E' 
cannot be the intersection of the two lines, since they 
do not touch; and it is also the same in regard to f f. 
Hence, when two right lines do not cut each other in 
space, the intersections of their projections are not in 
the same right line perpendicular to the common in¬ 
tersection of the planes of projection. 

VI. To find the angle made by a plane with the 
horizontal plane of projection. 

Let a b and a c, Fig. 9, be the horizontal and verti¬ 
cal traces of the given plane, i. e., the lines on which 
the given plane would, if produced, cut the horizontal 
and vertical planes of projection. Take any conven¬ 
ient point d in a b, and from it draw d e perpendicular 
to a b, and cutting the line of intersection x y in 
e, from e draw e d perpendicular to x y and cutting 
a c in d'. The angle made by the given plane with 
the horizontal plane of projection is such that, with 
a base d e, it has a vertical height e d'. Draw such an 


22 


MODERN CARPENTRY 


angle on the vertical plane of projection by setting ofl 
from e the distance e d" equal to e d, and joining d'd", 
The angle d'd" e is the angle required.* 



VII. The traces of a plane and the projections of 
a point being given, to draw through the point a 
plane parallel to the given plane. 

In the perspective representation (Figs. 10 and 11) 
suppose the problem solved, and let B C be the given 


*The angle made by a plane with the vertical plane of pro¬ 
jection can be found in a similar manner. If we imagine the 
part above x y in Fig. 9 to be the horizontal projection and 
the part below a? y to be the vertical projection—in other 
words, if Fig. 9 is turned upside down —a b becomes the 
vertical trace and a c the horizontal trace, and the angle d' 
d" e is the angle made by the given plane with the vertical 
plane of projection. 











SOLID GEOMETRY 


*23 


plane, and A C, A B its vertical and horizontal traces, 
and E F a plane parallel to the given plane, and G F, 
G E its traces. Through any point D, taken at pleasure 



I. Vertical and Horizontal Projections. II. Perspective View 
Fig. 10. 


on the plane E F, draw the vertical plane H J, the 
horizontal trace of which, I H, is parallel to G E. 
The plane H J cuts the plane E F in the line k 1', and 
its vertical trace GF in lh The horizontal projection 













24 


MODERN CARPENTRY 


of k'l'is II I, and its vertical projection k'l' ; and 
as the point D is in k T, its horizontal and vertical 
projections will be ~ and d'. Therefore, if through 
d be traced a line d 1, parallel to A B, that line will 
be the horizontal projection of a vertical plane passing 
through the original point D; and if an 1 be drawn 
the indefinite perpendicular L" T, and through d', 
the vertical projection of the given point, be drawn 
the horizontal line d' 1', cutting the perpendicular in 
!, then the line F G drawn through 1 parallel to A C, 
will be the vertical trace of the plane required; and 



the line G E drawn parallel to A B, its horizontal 
trace. Hence, all planes parallel to each other have 
their projections parallel, and reciprocally. In solving 
the problem, let A B, A C (Figs. 10, I) be the traces 
of the given plane, and d d the projections of the 
given point. Through d draw d 1 parallel to A B, 
and from I draw I Y perpendicular to A H. Join d 
d' and through d' draw d' 1' parallel to the line of in¬ 
tersection A H. Then F l'G drawn parallel to A C, 
and G E parallel to A B, are the traces of the requir* 
ed plane. 











SOLID GEOMETRY 


25 


VIII. The traces AB, B 0, and AD, DC, of two 
planes which cut each other being given, to find the 
projections of their intersections. 

The planes intersect each other in the straight line 
A C (Fig.II, 11), of which the points A and C are the 
traces, since in these points this line intersects the 
planes of projection. To find these projections, it is 
only necessary to let fall on the line of intersection in 
Fig. II, 1, the perpendiculars A a, C c, from the points 
A and C, and join A c, C a. Ac will be the horizontal 
projection, and C a the vertical projection of C A, 
which is the line of intersection or arris of the planes. 

IX. The traces of two intersecting planes being 
given, to find the angle which the planes make be¬ 
tween them. 

The angle formed by two planes is measured by that 
of two lines drawn from the same point in their inter¬ 
section (one along each of the planes), perpendicular 
to the line formed by the intersection. This will be 
better understood by drawing a straight line across 
the crease in a double sheet of note-paper at right 
angles to the crease; if the two leaves of the paper 
are then partly closed so as to form an angle, we have 
an angle formed by two planes, and this angle is the 
same as that formed by two lines which have been 
drawn perpendicular to the line of intersection of the 
two planes. These lines in effect determine a third 
plane perpendicular to the arris. If, therefore, the 
two planes are cut by a third plane at right angles to 
their intersection the solution of the problem is ob¬ 
tained. ’' 

On the arris A C (Fig. 12, 11) take at pleasure any 


MODERN CARPENTRY 




I. Vertical and Horizontal Projections. II. Perspective Vje? 
Fig. 12. 













SOLID GEOMETRY 


27 


point E, and suppose a plane passing through that 
point cutting the two given planes perpendicular to 
the arris. There results from the section a triangle 
D E F, inclined to the horizontal plane, and the angle 
of which, D E F, is the measure of the inclination of 
the two planes. The horizontal projection of that tri¬ 
angle is the triangle D e F, the base of which, F D, is 
perpendicular to A c (the horizontal projection of the 
arris AC), and cuts it in the point g, and the line E 
g is perpendicular to D F. The line g E is necessary 
perpendicular to the arris A C, as it is ,in the plane 
D E F, and its horizontal projection is g e. Now, sup¬ 
pose the triangle D E F turned on D F as an axis, and 
laid horizontally, its summit will then be at E", and 
D E" F is the angle sought. The perpendicular g E 
is also in the vertical triangle A C c, of which the 
arris is the hypothenuse and the sides A c, C C are 
the projections. This description introduces the so¬ 
lution of the problem. 

Through any point g (Fig. 12, 1) on the line Ac, 
the horizontal projection of the arris or line of inter¬ 
section of the two planes, draw F D perpendicular to 
A c; on A c (which for the moment must be consid¬ 
ered as a “line of intersection” or “ground line” for 
a second vertical projection) describe the vertical pro¬ 
jection of the arris by drawing the perpendicular 
c C, and then joining A C'. A C gives the true length 
and inclination of the arris. From g draw g E' per¬ 
pendicular to A C', and meeting it in E'; from E' let 
fall a perpendicular E e' or A c, meeting A c in e. 
F e D is the horizontal ^projection of the triangle F 
E D (see No. 11) and from this the vertical projection 


28 


MODERN CARPENTRY 


f' e' D can be drawn as shown. We have now ob¬ 
tained the vertical and horizontal projections of two 
intersecting lines, namely, F e, e D, and f' e', e' D, and 
by problem (4) the triangle which they make with each 
other can be found. It will be seen that g E' is the 



I. Horizontal Projection. II. Vertical Projection on Plane AD. 

III. Vertical Projection on Plane cj. 

Fig. 13. 

true altitude of the triangle f' e' D, as c E is equal to 
e' e. Set off therefore from g towards A a distance g 
E equal to g E; and join F E, E" D; F E" D is the angle 
made by the two intersecting planes. 













SOLID GEOMETRY 


2!) 



-I. Vertical and Horizontal Projections. II. Perspective Vi#*p 


Fig. 14. 


















30 


MODERN CARPENTRY 


X. Through a given point to draw a perpendicular 
to a given plane. 

Let a and a' (Fig. 14, II.) he the projections of the 
given point, and B C, C D the horizontal and vertical 
traces of the given plane. Suppose the problem solved, 
and that A E is the perpendicular drawn through the 
A to the plane B D, and that its* intersection with the 
plane is the point E. Suppose also a vertical plane 
a F to pass through A E', this plane would cut B D in 
the line g D, and its horizontal trace a h would be 
perpendicular to the trace B C. In the same way 
a' e', the vertical projection of A E, would be perpen¬ 
dicular to C D, the vertical trace of the plane B D. 
Thus we find that if a line a h is drawn from a, per¬ 
pendicular to B C, it will be the horizontal trace of 
the plane in which lies the required perpendicular 
A E, and h F will be the yertical trace of the same 
plane. From a', draw upon C D an indefinite perpen¬ 
dicular, and that line will contain the vertical pro¬ 
jection of A E, as a h contains its horizontal pro¬ 
jection. To find the point of intersection of the line 
A E with the given plane, construct the vertical pro¬ 
jection g D of the line of intersection of the two 
planes, and the point of intersection of that line with 
the right line draAvn through a', will be .the point 
sought. If from that point a perpendicular is let 
fall on a h, the point e will be the horizontal projec¬ 
tion of the point of intersection E. 

In Fig. 15, 1, let BC, C D be the traces of the given 
plane, and a, a' the projections of the given point. 
From the point a, draw a h perpendicular to B C; a 
h will be the horizontal projection of a plane passing 


SOLID GEOMETRY 


31 


vertically through a, and cutting the given plane. 
On a h as “ground line” draw a vertical projection 
as follows: From a, draw a A perpendicular to a g, 
and make it equal to a" a'; from h draw h D' per¬ 
pendicular to h a, and make h D' equal to h D; draw 
g D, which will be the section of the given plane by 



a vertical D g, and the angle h g D' will be the meas¬ 
ure of the inclination of the given plane with the hor¬ 
izontal plane; there is now to be drawn, perpendicu¬ 
lar to this line, a line A E through A, which will be 
the vertical projection qn a h of the line required. 
From the point of intersection E let fall upon aha 















32 


MODERN CARPENTRY 


perpendicular, which will give e as the horizontal pro¬ 
jection of E. Therefore a e is the horizontal projection 
of the required perpendicular, and a' e' its vertical 
projection on the original “ground line” h a". It 
follows from this problem that— Where a right line in 
space is perpendicular to a plane, the projections of 
that line are respectively perpendicular to the traces 
of the plane. 

XI. Through a given point to draw a plane perpen¬ 
dicular to a given right line. 

Let a, a' (Fig. 16, 1) he the projections of the given 
point A, and b c, b' c' the projections of the given 
line B C. 

The foregoing problem has shown that the traces of 
the plane sought must be perpendicular to the projec¬ 
tions of the line, and the solution of the problem 
consists in making to pass through A, a vertical plane 
A f (Fig. 16, 11), the horizontal projection of which 
will be perpendicular to b c. 

Through a (Fig. 16, 1) draw the projection a f 
perpendicular to b c. From f raise upon K L the in¬ 
definite perpendicular f f, which will be the vertical 
trace of the plane a f f perpendicular to the horizontal 
plane, and passing through the original point A (in 
No. 11). Then draw through a in the vertical pro¬ 
jection a horizontal line, cutting f f in f, which point 
should be in the trace of the plane sought; and as 
that plane must be perpendicular to the vertical pro¬ 
jection of the given right line draw through f a per¬ 
pendicular to b c, and produce it to cut K L in G. 
This point G is in the horizontal trace of the plane 
sought. All that remains therefore, is from G to draw 


SOLID GEOMETRY 


33 



Pig. 16. 

















34 


MODERN CARPENTRY 


G D perpendicular to b c. If the projections of the 
straight line are required, proceed as in the previous 
problem, and as shown by the dotted lines; the plane 
will cut the given line at k in the horizontal projec¬ 
tion, and at k in the vertical projection.* 

XII. A right line being given in projection, and 
also the traces of a given plane, to find the angle which 
the line makes with the plane. 

Let A B (Fig. 17, 11) be the original right line inter¬ 
secting the plane C E in the point B. If a vertical 
plane a B pass through the right line, it will cut the 
plane C E in the line f B, and the horizontal plane in 
the line a b. As the plane a B is in this case parallel 
to the vertical plane of projection, its projection on 
that plane will be a quadrilateral figure a b of the 
same dimensions, and f B contained in the rectangle 
will have for its vertical projection a right line D b, 
which will be equal and similar to f B. Hence the 
two angles a b D, A B f, being equal, will equally be 
the measure of the angle of inclination of the right 


♦The diagram will be less confusing if the projection on 
a h is drawn separately, as. in Fig. 16, where I is the hori¬ 
zontal projection or plan, II the vertical projection or eleva¬ 
tion on a plane parallel to ha", and III the vertical projection 
or elevation on a plane parallel to h a. To draw No. Ill draw 
first the ground-line h a equal to h a on No. I, and mark on it 
the point g; from h draw the vertical h d' equal to h d, and 
from a draw the vertical a a equal to a" af; join d' g and a h, 
and from the point of intersection e let fall a perpendicular on 
h a, cutting it in e. a e is the actual length and inclination 
of the required line, and a e its horizontal length. Transfer 
the length a e to No. I, and from e draw e e' perpendicular to 
h a" and cutting a' h in e'. If the drawing has been correctly 
made, a line from e parallel to the ground-line h a" will also 
intersect a' h in e'. 



SOLID GEOMETRY 


35 


line A B to the plane C E. Thus the angle a b D (Fig. 
17, 1), is the angle sought. 

This case presents no difficulty; but when the line 
is in a plane which is not parallel to the plane of 




Fig. 17. 


projection, the problem is more difficult; as, however, 
the second case is not of much practical value, it will 
not be considered. 















36 


MODERN CARPENTRY 


3. STRAIGHT-SIDED SOLIDS. 

XIII. Given the horizontal projection of a regular 
tetrahedron, to find its vertical projection. 

Let A B C d (Fig. 18) be the given projection of 
the tetrahedron, which has one of its faces coincident 
with the gronnd-line. From d draw an indefinite line 





Fig. 18. 


d D perpendicular to d C, and make C D equal to C B, 
C A, or A B; d D will be the height sought, which is 
carried to the vertical projection from c to d. Join 
A d and B d, and the vertical projection is complete. 
This problem might be solved in other ways. 








SOLID GEOMETRY 


37 


XIV. A point being given in one of the projections 
of a tetrahedron, to find the point on the other pro¬ 
jection. 

Let e be the point given in the horizontal projec¬ 
tion (Fig. 18). It may be considered as situated 
in the plane C B d, which is inclined to the horizontal 
plane, and of which the vertical projection is the tri¬ 
angle c B d. According to the general method, the 
vertical projection of the given point is to be found 
somewhere in a perpendicular raised on its horizontal 
projection e. If through d and the point e be drawn 
a line produced to the base of the triangle in f, the 
point e will be on that line, and its vertical projection 
will be on the vertical projection of that line fed, 
at the intersection of it with the perpendicular raised 
on e. If through e be drawn a straight line gh, par¬ 
allel to C B, this will be a horizontal line, whose ex¬ 
tremity h will be on B d. The vertical projection of 
d B is d B; therefore, by raising on h a perpendicular 
to A B, there will be obtained h, the extremity of a 
horizontal line represented by h g in the horizontal 
plane. If through h is drawn a horizontal line h g 
this line will cut the vertical line raised on e in e, 
the point sought. If the point had been given in g 
on the arris c d, the projection could not be found in 
the first manner; but it could be found in the second 
manner, by drawing through g, a line parallel to C B, 
and prolonging the horizontal line drawn through h, 
to the arris c d, which it would cut in g, the point 
sought. The point can also be found by laying down 
the right-angled triangle C d D (which is the devel¬ 
opment of the triangle formed by the horizontal pro- 


38 


MODERN CARPENTRY 


jection of the arris C d, the height of the solid, and 
the length of the arris as a hypotenuse), and by draw¬ 
ing through g the line g G perpendicular to C d, to in¬ 
tersect the hypotenuse in G, and carrying the height 
g G from c to g in the vertical projection. One or 
other of these means can be employed according to 
circumstances. If the point had been given in the 
vertical instead of the horizontal projection, the same 
operations inverted would require to be used. 



XV. Given a tetrahedron, and the trace of a plane 
(perpendicular to one of the planes of projection) cut¬ 
ting it, by which it is truncated, to find the projection 
of the section. 

First when the intersecting plane is perpendicular 
to the horizontal plane (Fig. 19), the plane cuts the 








SOLID GEOMETRY 


39 


base in two points ef, of which the vertical projec- 
tions are e and f; and the arris B d is cut in g, the 
vertical projection of which can readily be found in 
any of the ways detailed in the last problem. Having 
found g join egfg and the triangle egf is the pro¬ 
jection of the intersection sought. 


d' 



"When the intersecting plane is perpendicular to the 
vertical plane, as e f in Fig. 20, the horizontal projec¬ 
tions of the three points egf have to be found. The 
point g in this case may be obtained in several ways. 
First by drawing G g h through g, then through h 
drawing a perpendicular to the base, produced to the 
arris at h, in the horizontal projection, and then draw¬ 
ing h g parallel to C B> cutting the arris B d in g, 
which is the point required. Second, on dB, the hor- 










40 


MODERN CARPENTRY 


izontal projection of the arris, construct a triangle 
d D B, d B being the altitude of the tetrahedron, and 
B D the arris, and transfer this triangle to the ver¬ 
tical projection at d d B. From g draw the horizontal 
line cutting B d in G; g G is the horizontal distance 
of the required point in the arris, from the vertical 
axis of the tetrahedron as d is horizontal projection 
of the vertical axis, and dB the horizontal projection 
of the arris, it follows that the length gG, trans¬ 
ferred to d g, will give the required point' g. The 
points e and f are found by drawing lines from e and 
f perpendicular to the ground-line, and producing 
them till they meet the horizontal projections of the 
arrises in e and f. The triangle e f g is the horizontal 
projection of the section made by the plane e f. 

XVI. The projections of a tetrahedron being given, 
to find its projections when inclined to the horizontal 
plane in any degree. 

Let ABCd (Fig. 21) be the horizontal projection 
of a tetrahedron, with one of its sides coincident with 
the horizontal plane, and e d B its vertical projection; 
it is required to find its projections when turned round 
the arris AB as an axis. The base of the pyramid 
being a horizontal plane, its vertical projection is the 
right line c B. If this line is raised to c by turning 
on B, the horizontal projection will be A c2 B. When 
the point c, by the raising of B c, describes the arc 
c c, the point d will have moved to d, and the per¬ 
pendicular let fall from that point on the horizontal 
plane will give d3, the horizontal projection of the 
extremity of the arris C d; for as the summit d moves 
in the same plane as C, parallel to the vertical plane 


SOLID GEOMETRY 


41 


of projection the projection of the summit will evi¬ 
dently be in the prolongation of the arris C d, which 
is the horizontal projection or trace of that plane. 
The process therefore, is very simple and is as fol¬ 
lows: Construct at the point B the angle required, 
c B c, and make the triangle c d B equal to cdB; 



Fig. 21 . 


from d let fall a perpendicular cutting the prolonga¬ 
tion of the arris C d in d3, and from c a perpendicular 
cutting the same line in c2; join B c2, A c2, B d3, A d3. 

The following is a more general solution of the 
problem: Let ABCd (Fig. 22), be the horizontal 
projection of a pyramid resting with one of its sides 
on the horizontal plane, and let it be required to 
raise, by its angle C, the pyramid by turning around 
the arris A B, until its base makes with the horizontal 









42 


MODERN CARPENTRY 


plane any required angle, as 50°. Conceive the right 
line Ce turning round e, and still continuing to be 
perpendicular to A B, until it is raised to the required 
angle, as at e C. If a perpendicular be now let fall 
.from C, it will give the point C as the horizontal pro¬ 
jection of the angle C in its new position. Conceive 



a vertical plane to pass through the line C e. This 
plane will necessarily contain the required angle. Sup¬ 
pose, now we lay this plane down in the horizontal 
projection thus: Draw from e the line e C, making 





SOLID GEOMETRY 


43 


with e C an angle of 50°, and from e with the radius 
e c describe an arc cutting it in C. From C let fall 
on C e, a perpendicular on the point C, which will 
then be the horizontal projection of C in its raised 
position. On C e draw the profile of the tetrahedron 
C D e inclined to the horizontal plane. From D let 
fall a perpendicular on G e produced, and it will give 
d as the horizontal projection of the summit of the 
pyramid in its inclined position. Join Ad, B d, Ac, 
B c to complete the figure. The vertical projection of 
the tetrahedron in its original position is shown by a d 
b, and in its raised position by a, c2, d2, b, the points 
c2, and d2 being found by making the perpendiculars 
c3 c2 and d3 d2 equal to C C and d D respectively. 


H 



XVII. To construct vertical and horizontal pro¬ 
jections of a cube, the axis 1 of which are perpendic¬ 
ular to the horizontal plane. 

If an arris of the cube is given, it is easy to find its 
axis, as this is the hypotenuse of a right-angled tri¬ 
angle, the shortest side of which is the length of an 





44 


MODERN CARPENTRY 


arris, and the longest the diagonal of a side. Conceive 
the cube cut by a vertical plane passing through its 
diagonals EG, AC (Fig. 23), the section will be the 
rectangle A E G C. Divide this into two equal right- 
angled triangles, by the diagonal E C. If in the upper 
and lower faces of the cube, we draw the diagonals 
F H, B D, they will cut the former^ diagonals in the 
points f and b. Now, as the lines b B, b D, fF, fH, 
are perpendicular to the rectangular plane AEGC, 
f b may be considered as the vertical projection of 
BF and DH, and from this consideration we may 
solve the problem. 



Let AE (Fig. 24) be the arris of any cube. The 
letters here refer to the same parts as those of the 
preceding diagram (Fig. 23). Through A draw an 
indefinite line, A C, perpendicular to A E. Set off on 







SOLID GEOMETRY 


45 


this line, from A to C, the diagonal of the square of 
A E, and join E C, which is then the axis of the cube. 
Draw the lines EG, G G, parallel respectively to A C 
and A E, and the resulting rectangle, A E G C, is the 
section of a cube on the line of the diagonal of one of 
its faces. Divide the rectangle into two equal parts 
by the line b f, which is the vertical projection of 
the lines BP, DH (Fig. 23), and we obtain, in the 
figure thus completed, the vertical projection of the 
cube, as a c b d (Fig. 25). 

Through C (Fig. 24), the extremity of the diagonal 
E C draw y z perpendicular to it, and let this line 
represent the common section or ground-line of the 
two planes of projection. Then let us find the hor¬ 
izontal projection of a cube of which A E G C is the 
vertical projection. In the vertical projection the 
axis E C is perpendicular to y z, and consequently, 
to the horizontal plane of projection, and we have 
the height above this plane of each of the points which 
terminate the angles. Let fall from each of these 
points perpendiculars to the horizontal plane, the 
projections of the points will be found on these per¬ 
pendiculars. The horizontal projection of the axis E C 
will be a point on its prolongation, as c. This point 
might have been named e with equal correctness, as it 
is the horizontal projection of both the extremities of 
the axis, C and E. Through c draw a line parallel 
to yz, and find on it the projections of the points A 
and G, by continuing the perpendiculars A a, G g, to 
a and g. We have now to find the projections of the 
points b f (representing D B F LI, Fig. 23,), which will 
be somewhere on the perpendiculars b b, ff, let fall 


46 


MODERN CARPENTRY 


from them. We have seen in Fig. 23 that B F, D H 
are distant from b f by an extent equal half the diag¬ 
onal of the square face of the cube. Set off, therefore, 
on the perpendiculars b b and f f, from o and m, the 
distance. A b in d, b, and f, f and join da, a b, b f , 
f g, g f, to complete the hexagon which is the hori¬ 
zontal projection of the cube. Join f e, f e, and a c, to 
give the arrises of the upper half of the cube. The 
dotted lines d c, be, g c, show the arrises of the lower 
side. Knowing the heights of the points in these ver¬ 
tical projections, it is easy to construct a vertical pro¬ 
jection on any line whatever, as that on R S below. 



d 



XVIII. To construct the projections of a regular 
octahedron, when one of its axis is perpendicular to 
either plane of projection. 

Describe a circle (Fig. 26), and divide it into four 
equal parts by the diameters, and draw the lines a d, 
db, be, ca; a figure is produced which serves for 
either the vertical or the horizontal projection of the 
octahedron, when one of its axis is perpendicular to 
either plane. 












SOLID GEOMETRY 


47 


XIX. One of the faces of an octahedron being given, 
coincident with the horizontal plane of projection, to 
construct the projections of the solid. 

Let the triangle ABC (Fig. 27) be the given face. 
If A be considered to be the summit of one of the two 
pyramids which compose the solid, B C will be one of 
the sides of the square base, k C B i. The base makes 
with the horizontal plane an angle, which is easily 
found. Let fall from A a a perpendicular on B c, cut¬ 
ting it in d, with the length B c as a radius, and from 


H 



Fig. 27. 


d as a centre, describe the indefinite arc e f. The per¬ 
pendicular A d will be the height of each of the faces, 
and, consequently, of that which, turning on A, should 
meet the side of the base which has already turned on d. 
Make this height turn on A, describing from that point 
as a centre, with the radius A d, an indefinite arc, cut¬ 
ting the first arc in G, the point of meeting of one of 
the faces with the square base; draw the line G A, G d: 
the first is the profile or inclination of one of the faces 
on the given face ABC according to the angle d A G; 
the second, d G, is the iiiclination of the square base, 




48 


MODERN CARPENTRY 


which separates the two pyramids in the angle A d G. 
The face adjacent to the side B C is found in the same 
manner. Through G, draw the horizontal line G H 
equal to the perpendicular A d. This line will be the 
profile of the superior face. Draw dH, which is the 
profile of the face adjacent to B C. From H let fall a 
perpendicular on A d produced, which gives the point 
h for the horizontal projection of H, or the summit of 
the superior triangle parallel to the first, draw h i par¬ 
allel to CA, hk parallel to AB, Ak and Bh per¬ 
pendicular to A B, and join k C, C h, B i, and A i and 
the horizontal projection is complete. From the heights 
we have thus obtained we can now draw the vertical 
projection shown in No. II, in which the parts have the 
same letters of reference. 

The finding of the horizontal projection may be 
abridged by constructing a hexagon and inscribing in 
it the two triangles A C B, h i k. 

XX. Given in the horizontal plane the projection of 
one of the faces of a dodecahedron, to construct its 
projections. 

The dodecahedron is a twelve-sided solid, all the 
sides being regular and equal pentagons. It is neces¬ 
sary, in order to construct the projection, to discover 
the inclination of the faces among themselves. Let 
the pentagon ABODE (Fig. 28) be the side on which 
the body is supposed to be seated on the plane. Con¬ 
ceive two other faces, EFGHD and DIKLC, also 
jn the horizontal plane, and then raised by being 
turned on their bases, ED, DC. By their movement 
they will describe in space arcs of circles, which will 
terminate by the meeting of the sides D H, D I. 


SOLID GEOMETRY 


49 


To find the inclination of these two faces.—From the 
points 1 and H let fall perpendiculars on their bases 
produced. If each of these pentagons were raised ver¬ 
tically on its base, the horizontal projection of H and 1 
would be respectively in z z • but as both are raised 



together, the angles H and I would meet in space above 
h where the perpendiculars intersect therefore, h will 
be the horizontal projection of the point of meeting of 
the angles. To find the horizontal projection of K, pro¬ 
long indefinitely z I, and set off from z on z I the length 
xK in k; from z as centre with radius z I, describe 












50 


MODERN CARPENTRY 


the arc II cutting the perpendicular hi in I; jo.n zl; 
then from z as centre, with the radius z k, describe an 
arc cutting z I produced in the point K, from which 
let fall on z k a perpendicular K k, and produce it to 
k in x K. If, now, the right-angled triangle z k K, were 
raised on its base, k would be the projection of K. 
Conceive now the pentagon C D I K L turned round on 
C D, until it makes an angle equal to k z K with the 
horizontal plane, the summit K will then be raised 
above k by the height k K, and will have for its hori¬ 
zontal projection the point k. In completing the figure 
practically ;—from the centre o, describe two concen¬ 
tric circles passing through points h D. Draw the 
lines h D, h k, and carry the last round the circum¬ 
ference in mnoprstu: through each of these lines 
draw radially the lines m C, o B, r A, t E, and these 
lines will be the arrises analogous to h D. This being 
done, the inferior half of the solid is projected. By 
reason of the regularity of the figure, it is easy to see 
that the six other faces will be similar in those already 
drawn, only that although the superior pentagon will 
have its angles on the same circumference as the in¬ 
ferior pentagon the angles of the one will be in the 
middle of the faces of the other. Therefore, to de¬ 
scribe the superior halfthrough the angles npsvk, 
draw the radial lines n 1, p 2, s 3, v4, k 5, and join 
them by the straight lines 1 2, 2 3, 3 4, 4 5, and 5 1. 

I o obtain the length of the axis of the solid, observe 
that the point k is elevated above the horizontal plane 
by the height k K: carry that height to k K: the point 
r, analogous to h, is raised the same height as that 
point, that is to say h i, which is to be carried from r 


SOLID GEOMETRY 


51 


to R; and the line R K is the length sought. As this 
axis should pass through the centre of the body, if a 
Vertical projection of the axis in 0, and therefore 0 o 
I‘s the half of the height of the solid vertically. By 
doubting this height, and drawing a horizontal line to 
cut the vertical lines of the angles of the superior face 
is obtained, as in the upper portion of (Fig. 28), in 
which the same letters refer to the same parts. 

XXI. One of the faces of a dodecahedron being 
given, to construct the projections of the solid, so that 
its axis may be perpendicular to the horizontal plane. 

Let ABEDC (Fig. 29, 1) be the given face. The 
solid angles of the dodecahedron are each formed by 
the meeting of three pentagonal planes. If there be 
conceived a plane B C passing through the extremities 
of the arrises of the solid angle A, the result of the 
section would be a triangular pyramid, the sides of 
whose base would be equal to one of the diagonals of 
the face, such as B C. An equilateral triangle b c f 
(Fig. 29, 11) will represent the base of that pyramid 
inverted, that is, with its summit resting on the hori¬ 
zontal projection, it it required to find the height of that 
pyramid, or which is the same thing, that one of the 
three points of its base b c f, for as they are all equally 
elevated, the height of one of them gives the others. 
There is necessarily a proportion between the triangle 
Abe (No. 11) and ABC (No. 1), since the first is the 
horizontal projection of the second. A g is the hori¬ 
zontal projection of A G; but A G is a part of A H, and 
the projection of that line is required for one of the 
faces of the solid; therefore as AG:Ag::AH:x. In 
other words, the length, the length of x may be ob- 


52 


MODERN CARPENTRY 


tained by drawing a fourth proportion of the three 
lines it will be found to be equal to A h: or it may be 
obtained graphically thus:—Raise on A g at g an in¬ 
definite perpendicular, take the length AG (No. 1) 
and carry it from A to G (No. 11); g is a point in the 


a" 



assumed pyramidal base b c f. Since A G is a portion 
of A H, A G will be so also. Produce A G, therefore, 
to H, making AH equal to AH (No. 1) and from H 
let fall a perpendicular on A g produced, which gives 
h the point sought. Produce H h, and carry on it the 














SOLID GEOMETRY 


53 


length HD or HE from h to d and h to e; draw the 
lines c d, be, and the horizontal projection of one of 
the faces is obtained inclined to the horizontal plane, in 
the angle HAh. As the other two inferior faces are 
similar to the one found, the three faces should be 
found on the circumference of a circle traced from A 
as centre, and with A d or A e as a radius. Join f A, 
prolong An, Ao, perpendicular to the sides of the 
triangle f c b, and make them equal to A h, and through 
their extremities draw perpendiculars, cutting the cir¬ 
cumference in the points ik, lm. Draw the lines ib, 
k f, 1 f, m c, and the horizontal projections of the three 
inferior faces obtained. The superior pyramid is sim¬ 
ilar and equal to the inferior, and solely opposed by its 
angles. Describe a circle passing through the three 
points of the first triangle, and draw within it a second 
equilateral triangle nop, of which the summits corres¬ 
pond to the middle of the faces of the former one. 
Each of these points will be the summit of a pentagon, 
as the points b c f. These pentagons have all their 
sides common, and it is only necessary therefore to 
determine one of these superior pentagons to have all 
the others. Six of the faces of the dodecahedron have 
now been projected; the remaining six are obtained by 
joining the angular points already found, as qd, e r, 
t i, k s, &c. 

To obtain the vertical projection (No. Ill) begin 
with the three inferior faces. The point A in the 
horizontal projection being the summit of the inferior 
solid angle, will have its vertical in a; the points b c f, 
when raised to the height g G, will be in b c f, or simply 
b f. The points b g c being in a plane perpendicular to 


54 


MODERN CARPENTRY 


the vertical plane, will necessarily have the same ver¬ 
tical projection, b. Tlie line a f will be the projection 
of the arris A f, and a b will be that of the arrises A b, 
A c, and of the line A g, or rather that of the triangle 
Abe, which is in a plane perpendicular to the vertical 
plane. But this triangle is only a portion of the given 
pentagonal face (No. 1), of which AH is the perpen¬ 
dicular let fall from A on the side E D. Produce a b 
to e, making a e equal to AH; e is the vertical projec¬ 
tion of the arris e d. This arris is common to the in¬ 
ferior pentagon, and to the superior pentagon e d q p r, 
which is also perpendicular to the vertical plane, and, 
consequently, its vertical projection will be e p, equal 
to a e. 

This projection can now be obtained by raising a 
vertical line through p, the summit of the superior 
pentagon, and from e as a centre, and with the radius 
AH or AH, describing an arc cutting this line in p, 
the point sought. But p n o belong to the base of the 
superior pyramid; ytherefore, if a perpendicular is 
drawn from n through yzton, and the height p is 
transferred to n by drawing through p' a line parallel 
to y z, n will be the projection of the points n and o. 
Through n draw sn a parallel to a e cutting perpen¬ 
diculars drawn through s and A in the horizontal pro¬ 
jection. Through s draw sf parallel to p e, and join 
a f, a p; set off on the perpendicular from r the height 
of s above y z at r, and draw r t parallel to y z, cutting 
the perpendicular from t, and joint nt. Draw perpen¬ 
diculars from k and i through y z, to k and i, make k 
and i the same height as e, and draw k i, and join i b, 
i t. The vertical projection is now complete. 


SOLID GEOMETRY 


55 


XXII. In a given sphere to inscribe a tetrahedron, a 
hexahedron or cube, an octahedron and a dodecahed¬ 
ron. 

Let AB (Fig. 30) be the diameter of the given 
sphere. Divide it into three equal parts, D B being one 
of these parts. Draw DE perpendicular to AB, and 
draw the chords A E, E B. A E is the arris of the tetra- 


H 



hedron, and E B the arris of the hexahedron or cube. 
From the centre C draw the perpendicular radius CF, 
and the chord F B is the arris of the octahedron. Divide 
B E in extreme and mean proportion in G, and B G 
is the arris of the dodecahedron. The arrises being 
known, the solids can be drawn by the help of the prob¬ 
lems already solved. Draw the tangent AII equal to 
AB; join HC and A I; A I is the arris of an icosahe- 







56 


MODERN CARPENTRY 


dron which can be inscribed in the sphere, an icosa¬ 
hedron being a solid with twenty equal sides, all of 
which are equilateral triangles. See Fig. 30 V 2 . 



Fig. 30% 











SOLID GEOMETRY 


57 


4. THE CYLINDER, CONE, AND SPHERE. 

XXIII. The horizontal projection of the cylinder, 
the axis of which is perpendicular to the horizontal 
plane, being given to find the vertical projection. 

Let the circle ABCD (Fig. 31) be the base of the 
cylinder, and also its horizontal projection. From the 
points A and C raise perpendiculars to the ground-line 



B 

Fig. 31. 


a c, and produce them to the height of the cylinder— 
say, for example, a e, c f. Draw e f parallel to a c, and 
the rectangle a e f c is the vertical projection required. 

XXIV. Given the traces of an oblique plane, to de¬ 
termine the inclination of the plane to both the H. P. 
and the V. P. 










58 


MODERN CARPENTRY 


Let y t and h t (Fig. 32) be the traces of the given 
plane. Draw the projections of a semi-cone having its 
axis a'b' in the vertical plane, the apex a' in the given 
vt and its base (a semi-circle) ced in the hp and 
lying tangentially to the given h t. Then the base angle 
(0) of the cone gives the inclination of the plane to the 
hp. To determine the inclination of the plane to the 
v. p., draw the projections of a second semi-cone, hav¬ 
ing the axis mn in the h p, and the apex m in the 



given h t, while the base is in the v p and tangential 
to the v t. The base angle (0) of this cone gives the 
inclination to the v p. 

XXV. The base of a cylinder being given, and also 
the angles which the base makes with the planes 
of projection, to construct the projections of the cyl¬ 
inder. 

Let the circle AGBH (Fig. 33) be the given base,and 
let each of the given angles be 45°. Draw the diameter 
AB,makinganangle of 45° with the ground line or ver- 



SOLID GEOMETRY 


59 


tical plane, and draw the line A B, making with A B the 
given angle; and from A as a centre, w T ith A B and 
A C as radii, describe arcs cutting A B in B and C. 
Then draw A D, B E perpendicular to A B and equal 



\ 



Fig. 33. 


to the length of the cylinder; the rectangle A E is 
the vertical projection of the cylinder parallel to the 
vertical plane and inclined to the horizontal plane 
A B in an angle of 45°. Now prolong indefinitely 
the diameter B A, and this line will represent the 
projection on the horizontal plane of the line in which 
the generating circle moves to produce the cylinder. 








60 


MODERN CARPENTRY 


If from B and C perpendiculars be let fall on A B, 
k will be the horizontal projection of B, A k that of 
the diameter A B, and c that of the centre C. Through 
c draw h g perpendicular to A B, and make c h, c g 
equal to C H, G G; and the two diameters of the 
ellipse, which is the projection of the base of the cyl¬ 
inder, will be obtained; namely, A k and h g. 

In like manner, draw D F E the lines D d, F f, E c, 
perpendicular to the diameter A B produced, and 
their intersections with the diameter and the sides of 
the cylinder will give the means of drawing the 
ellipse which forms the projection of the farther 
end of the cylinder. The ellipses may also be found 
by taking any number of points in the generating 
circle as I J, and obtaining their projections i j. The 
method of doing this, and also of drawing the verti¬ 
cal projection c f, will be understood without fur¬ 
ther explanation. 

XXVI. A point in one of the projections of a cone 
being given, to find it in the other projection. 

Let a (Fig. 34) be the given point. This point be¬ 
longs equally to the circle which is a section of the 
cone by a plane passing through the point parallel to 
the base, and to a straight line forming one of the 
sides of a triangle which is the section of the cone by a 
plane perpendicular to its base and passing through 
its vertex and through the given point, and of which 
f a g is the horizontal, and f a g the vertical projec¬ 
tion. To find the vertical projection of a, through a 
draw a a perpendicular to b c, and its intersection with 
f g is the point required; and reciprocally, a in the 


SOLID GEOMETRY 


61 


horizontal projection may be found from a in the 
vertical projection, in the same manner. 

Otherwise, through a, in the horizontal projection, 
describe the circle a d c, and draw e e or c c, cutting 



the sides of the cone in e and c; draw c e parallel 
to the base, and draw a a, cutting it in a, the point 
required. 








62 


MODERN CARPENTRY 


XXVII. On a given cylinder to describe a helix. 

Let abed, &c. (Fig. 35), be the horizontal projec¬ 
tion of the given cylinder. Take on this curve a series 
of equal distances, a b, b c c d, &c., and through each 


n 



of the points a, b, c, &c. draw a vertical line, and pro¬ 
duce it along the vertical projection of the cylinder. 
Then conceive a curve cutting all these verticals in the 
points abed, in such a manner that the height of the 
point above the ground-line may be in constant rela¬ 
tion to the arcs a b, b c, c d; for example, that a may 

















SOLID GEOMETRY 


63 


be the zero of height, that bb may be 1, c c 2, d d 3, 
&c.; then this curve is named a helix. To construct 
this curve, carry on the vertical projection on each 
vertical line such a height as has been determined, as 
1 on b, 2 on c, 3 on d; and through these points will 
pass the curve sought. It is easy to see that the curve 



so traced is independent of the cylinder on which 
it has been supposed to be traced; and that if it be 
isolated, its horizontal projection will be a circle. The 
helix is named after the curve which is its horizontal 
projection. Thus the helix in the example is a helix 
with a circular base. The vertical line f n is the axis 


























64 


MODERN CARPENTRY 


of the helix, and the height b b, comprised between 
two consecutive intersections of the curve with a verti¬ 
cal, is the pitch of the helix. 

XXVIII. On a given cone to describe a helix. 

Let the projections of the given cone be as shown in 
Fig. 36. Divide the base of the cone in the horizon¬ 
tal projection into any number of equal parts, as a b, 
be, c d, &c., and draw lines from the vertex to the 
points thus obtained. Set off along these lines a series 
of distances increasing in constant ratio, as 1 at b, 2 
at c, 3 at d, &c. The curve then drawn through these 
points when supposed to be in the same plane, is called 
a spiral. If these points, in addition to approaching 
the centre in a constant ratio, are supposed also to 
rise above each other by a constant increase of height, 
a helical curve will be obtained on the vertical pro¬ 
jection of the cone.* 

XXIX. A point in one of the projections of the 
sphere being given, to find it in the other projection. 

Let a be the given point in the horizontal projection 
of the sphere hb c i (Fig. 38). Any point on the sur¬ 
face of a sphere belongs to a circle of that sphere. 
Therefore, if a is a point, and a vertical plane b c is 
made to pass through that point to A B, the section of 


*The octahedron is formed by the union of eight equilateral 
triangles; or, more correctly, by the union of two pyramids 
with square bases, opposed base to base, and of which all the 
solid angles touch a sphere in which they may be inscribed. 

It is essential that the diagram should be clearly seen as a 
solid, and not as a mere set of lines in one plane. Imagine h 
as the apex of one pyramid on the base Tc c b i, and a as the 
apex of the other pyramid on the opposite side of the same 
base. The octahedron is shown to be lying on its side a b c. 



SOLID GEOMETRY 


65 


the sphere by this plane will be a circle, whose diam¬ 
eter will be b c, and the radius consequently, d b or 
d c; and the point a will necessarily be in the circum¬ 
ference of this circle. Since the centre d of this circle 
is situated on the horizontal axis of the sphere, and 




as this axis is perpendicular to the vertical plane, its 
vertical projection will he the point d. It is evident 
that the vertical projection of the given point a will 
be found in the circumference of the circle described 
from d with the radius d b or d c and at that point 
of it where it is intersected by the line drawn through 



























66 


MODERN CARPENTRY 


a, perpendicular to A B. Its vertical projection will 
therefore be either a or a, according as the point a is 
on the superior or inferior semi-surface of the sphere. 

The projection of the point may also be found thus: 
Conceive the sphere cut by a plane parallel to the hori¬ 
zontal plane of projection passing through the given 
point a. The resulting section will be the horizontal 
circle described from k, with the radius k a; and the 
vertical projection of this section will be the straight 
line g e, or g e; and the intersections of these lines 
with the perpendicular drawn through a, will be the 
projection of a, as before. 


5. SECTIONS OF SOLIDS. 

To draw sections of any solid requires little more 
than the application of the method described in the 
foregoing problems. Innumerable examples might be 
given, but a few selected ones will suffice. 

XXX. The projections of a regular tetrahedron be- 
ing given, to draw the section made by a plane perpen¬ 
dicular to the vertical plane and inclined to the hori¬ 
zontal plane. 

Let A B C D and abed (Fig. 39) be the given pro¬ 
jections, and E F G the given plane perpendicular to 
the vertical plane and inclined to the horizontal plane 
at an angle of 30°. The horizontal projection ef g of 
the section is easily found as shown. To find the cor¬ 
rect section draw through e, f, and g, lines parallel to 



SOLID GEOMETRY 


67 


the ground-line A C, and a c one of them as E e set 
off the distances E F, E G at E f and E g, and through 
f and g draw perpendiculars cutting the other lines in 
F and G'. Join E G, G F, and F E. E F G is the cor¬ 
rect section made by the plane. 



XXXI. The projections of a hexagonal pyramid be¬ 
ing given, to draw the section made by a plane perpen¬ 
dicular to the vertical plane and inclined to the hori¬ 
zontal plane. 

Let A B C D and a b c d e f g (Fig. 40) be the given 
projections, and H I J K the given plane. The hori- 













68 


MODERN CARPENTRY 


zontal projection h i j k 1 m of the section is easily 
found as shown. Through m, 1, h, j, and i draw lines 
parallel to the ground-line A D, and on one of them 
as h m, set off the distances H I, H J, H K, at h M, 



th 1, h k. From h, M, 1, and k draw perpendiculars 
meeting the other lines in H, I, L, J, and K, and join 
the points of intersection. H I J K L M is the true 
section made by the plane. 






















SOLID GEOMETRY 


69 


XXXII. The projections of an octagonal pyramid 
being given, to draw the section made by a verti¬ 
cal plane. 

Let A B C D F and a b c d e f (Fig. 41) be the given 
projections, and g h i j k of the section made by the 
plane is easily found by drawing g g, h h, i i, &c., perpen¬ 



dicular to the ground-line AD. On A D produced set 
off the distances g h, h i, i j, and j k at G h, h i, &c., 
and from the points thus found draw perpendiculars to 
G K meeting lines drawn from h, i, and j parallel to 
A D, in H, I, and J. Join G Ef, H I, I J, and J K. 
G III J K is the true section made by the plane. 
























70 


MODERN CARPENTRY 


A cylinder may be cut by a plane in three different 
ways—1st, the plane may be parallel to the axis; 2nd, 
it may be parallel to the base; 3rd, it may be oblique 
to the axis or the base. 

In the first case the section is a parallelogram, whose 
length will be equal to the length of the cylinder, and 
whose width will be equal to the chord of the circle of 
the base in the line of section. Whence it follows, that 
the largest section of this kind will be that made by a 
plane passing through the axis; and the smallest will 
be when the section plane is a tangent—the section in 
that case will be a straight line. 

When the section plane is parallel to the base, the 
section will be a circle ^equal to the base. When the 
section plane is oblique to the axis or the base, the 
section will be an ellipse. 



XXXIII. To draw the section of a cylinder by a 
plane oblique to the axis. 

Let A B C D (Fig. 42) be the projection of a cylin¬ 
der, of which the circle EHFK represents the base 
divided into twenty equal parts at b c d e, &c., and let 
it be required to draw the section made by the plane 




















SOLID GEOMETRY 


71 


A C. The circular base must be drawn in such a posi¬ 
tion that the axis of the cylinder when produced 
meets the centre of the circle. Through the centre of 
the circle draw the diameter H K perpendicular to the 
axis produced. Then through the divisions of the 
base, bed, &c., draw lines parallel to the axis, and 
meeting the section plane in 1, 2, 3, &c., and through 
these points draw perpendiculars to A C making them 
equal to the corresponding perpendiculars from H K, 
i e, 1 b, 2 c, 3 d, &c. A curve drawn through the points 
thus found will be an ellipse, the true section of 
A B C D on the plane A C. # 

The Cone. A cone may be cut by a plane in five dif¬ 
ferent ways, producing what are called the conic sec¬ 
tions: 1st. If it is cut by a plane passing through the 
axis, the section is a triangle, having the axis of a 
cone as its height, the diameter of the base for its base, 
and the sides for its sides. If the plane passes through 
the vertex, without passing through the axis, as c e 
(Fig. 43), the section will still be a triangle, having 
for its base the chord c e, for its altitude the line c e, 
and for its sides the sides of the cone, of which the 
lines c e, o e are the horizontal and tho line c e the 
vertical projections. 2nd. If the cone is cut parallel to 
base, as in g h, the section will be a circle, of which 
g h will be the diameter. 3rd. When the section plane 
is oblique to the axis, and passes through the oppo¬ 
site sides of the cone, as m p h, the section will be 
an ellipse, m n h. 4th. When the plane is parallel to 
one of the sides of the cone, as r h, the resulting sec- 


*The point is assumed to be in the inferior half of the 
cylinder. 



72 


MODERN CARPENTRY 



tion is a parabola r s h t u. 5th. When the section 
plane is such as to pass through the sides of another 
cone formed by producing the sides of the first be- 
























SOLID GEOMETRY 


73 


yond the vertex, as the plane q h, the resulting curve 
in each cone in a hyperbola. 

Several methods of drawing the curves of the conic 
sections have already been given in Plane Geometry, 
Yol. 1. Here their projections, as resulting from the 
sections of the solid by planes, are to be considered. 
If the mode of finding the projections of a point on the 
surface of a given cone be understood, the projections 
of the curves of the conic sections will offer no diffi¬ 
culty. Let the problem be: First, to find the projec¬ 
tions of the section made by the plane m h. Take 
at pleasure upon the plane the several points, as p, &c. 
Let fall from these points perpendiculars to the hori¬ 
zontal plane, and on these will be found the horizontal 
projection of the points; thus, in regard to the point p 

-. Draw through p a line parallel to A B: this line 

will be the vertical projection of the horizontal plane 
cutting the cone, and its horizontal projection will be 
a circle, with s n for its radius. With this radius, 
therefore, from the centre e, describe a circle cutting, 
twice, the perpendicular let fall from p; the points of 
intersection will be two points in the horizontal pro¬ 
jection of the circumference of the ellipse. In the 
same manner, any other points may be obtained in 
its circumference. The operation may often be 
abridged by taking the point p in the middle of the 
line m h; for then m h will be the horizontal pro¬ 
jection of the major axis, and the two points found 
on the perpendicular let fall from the central point 
p will give the minor axis. 

To obtain the projections of the parabola, more 
points are required, such as r, 2, s, 3, h, but the mode 



74 


MODERN CARPENTRY 


of procedure is the same as for the ellipse. The ver¬ 
tical projection of the parabola u s h t u is shown at 
n s h t u. 

The projections of the section plane which produces 
the hyperbola are in this case straight lines, q h, z h. 

XXXIV. To draw the section of a cone made by a 
plane cutting both its sides, i. e., an ellipse. 



Let ADB (Fig. 44) be the vertical projection of 
the cone ACB the horizontal projection of half its 
base, and E F the line of section. From the points E 
and F let fall on AB the perpendiculars EG, FH. 





























SOLID GEOMETRY 


75 


Take any points in E F, as k, 1, m, n, &c., and from 
them draw lines parallel to A B, as k p, 1 q, m r, &c., 
and also lines perpendicular to A B, as k 1, 12 m 3, &c. 
Also from p, q, r, &c., let fall perpendiculars on AB, 
namely, pa, q z, r y, &c. From the centre of the base 
of the cone, 1, with radius la, lz, ly, &c., describe arcs 
cutting the perpendiculars let fall from k, 1, m, &c., in 

l, 2, 3, &c. A curve traced through these points will 
be the horizontal projection of the section made by the 
plane E F. To find the true section,—Through k, 1, 

m, &c., draw kt, lu, m v, n w, perpendicular through 
E F, and make them respectively equal to the corre¬ 
sponding ordinates, 5 1, 6 2, 7 3, &c., of the horizontal 
projection G 4 H, and points will be obtained through 
which the half E w F of the required ellipse can be 
traced. It is obvious that, practically it is necessary 
only to find the minor axis of the ellipse, the major 
axis EF being given. 

XXXV. To draw the section of a cone made by a 
plane parallel to one of its sides, i. e., a parabola. 

Let ADB (Fig. 45) be the vertical projection of a 
right cone, and A C B half the plan of its base; and 
let E F be the line of section. In E F take any number 
of points, E, a, b, c, e, F, and through them draw lines 
EH, a 61, b 7 2, &c., perpendicular to AB, and also 
lines parallel to AB, meeting the side of the cone in 
f, g, h, k, 1: from these let fall perpendiculars on A B, 
meeting it in m n, o, p, q. From the centre of the base 
1, with the radii lm, In, lo, &c., describe arcs cutting 
the perpendiculars let fall from the section line in the 
points 1, 2, 3, 4, 5; and through the points of intersec¬ 
tion trace the line H12345G, which is the horizontal 


76 


MODERN CARPENTRY 


projection of the section. To find the true section, 
from E, a, b, c, d, e, raise perpendiculars to EP, and 
make them respectively equal to the ordinates in the 
horizontal projection, as E r equal to E H, as equal to 


r> 



6 1, &c., and the points rstuvw in the curve will be 
obtained. The other half of the parabola can be drawn 
by producing the ordinates we, v d, &c., and setting 
the same distances to the right of E F. 



















SOLID GEOMETRY 


77 


XXXVI. To draw the section of a cone made by a 
plane parallel to the axis, i. e., an hyperbola. 

Let d c d (Fig. 46) be the vertical projection of the 
cone, d q r d one half of the horizontal projection of 
the base, and q r the section plane. Divide the line r q 
into any number of equal parts in 1, 2, 3, h, &c., and 


c 



Fig. 46. 


through them draw lines perpendicular to d d. From 
c as centre, with the radii cl, c2, &c., describe the arcs 
cutting d d; and from the points of intersection draw 
perpendiculars cutting the sides of the cone in 1, 2, 3, 
and these heights transferred to the corresponding per¬ 
pendiculars drawn directly from the points 1, 2, 3, &c., 
in r q, will give points in the curve. 
















73 


MODERN CARPENTRY 


XXXVII. To draw the section of a cuneoid made 
by a plane cutting- both its sides. 

Let A C B (No. 1, Fig. 47) be the vertical projection 
of the cuneoid, and A 5 B the plan of its base, and 
AB (No. 4) the length of the arris at C, and let DE 
be the line of section. Divide the semi-circle of the 
base into any number of parts, 1, 2, 3, 4, 5, &c., and 
through them draw perpendiculars to AB, cutting it 
in 1, m, n, o, p, &c., and join Cl, C m, Cn, &c., by lines 
cutting the section line in 6, 7, 8, 9 &c. From these 
points draw lines perpendicular to D E and make 
them equal to the corresponding ordinates of the semi¬ 
circle, either by transferring the lengths by the com¬ 
passes, or by proceeding as shown in the figure. The 
curve drawn through the points thus obtained will give 
the required section. 

The section on the line D K is shown in No. 2, in 
which A B equals D K; and the divisions e f g h k in 
D K, &c., are transferred to the corresponding points 
on AB; and the ordinates el, f m, gn, &c., are made 
equal to the corresponding ordinates 11, m2, n3, of the 
semi-circle of the base. In like manner, the section of 
the line GH, shown in No. 3, is drawn.* 

XXXVIII. To describe the section of a cylinder 
made by a curve cutting the cylinder. 

Let A B D E (Fig. 48) be the projection of the cylin¬ 
der, and C D the line of the section required. On A B 


*A cuneoid is a solid ending in a straight line, in which, if 
any point be taken, a perpendicular from that point may’be 
made to coincide with the surface. The base of the cuneoid 
may be of any form; but in architecture it is usually semi¬ 
circular or semi-elliptical, and parallel to the straight line 
forming the other end. 



SOLID GEOMETRY 


70 


describe a semi-circle, and divide it into any number of 
parts. From the points of division draw ordinates 1 h, 
2k, 31, 4m, &c., and produce them to meet the line of 
section in o, p, q, r, s, t, u, v, w. Bend a rule or slip 
of paper to the line C D, and prick off on its points C, 



o, p, q, &c.; then draw any straight line F G, and, un¬ 
bending the rule, transfer the points C, o, p, q, &c., to 
F, a, b, c, d, &c. Draw the ordinates al, b 2, c 3, &c., 
and make them respectively equal to the ordinates h 1, 
k2, 13, &c., and through the points found trace the 
curve. 







































80 


MODERN CARPENTRY 


XXXIX. To describe the section of a sphere. 

Let A B D C (Pig. 49) be the great circle of a sphere, 
and P G the line of the section required. Then since all 
the sections of a globe or sphere are circles, on P G de¬ 
scribe a semi-circle F 4G, which will be the section re¬ 
quired. 



Or, in P G take any number of points, as m, 1, k, H, 
and from the centre of the great circle E, describe’the 
arcs H n, k o, 1 p, m q, and draw the ordinates H 4, k 3, 
12, ml, and n4, o3, p2, q 1; then make the ordinates 
on P G equal to those on B C, and the points so ob¬ 
tained will give the section required.* 


*The projections of sections of spheres are, if the section 
panes are oblique, either straight lines or ellipses, and are 
round as follows: 

Let a & (Fig. 50 ) be the horizontal projection of the sec- 

P i a ? e * ? n f>1 the h ? e of section take any number of points 
as a c b, and through each of them draw a line perpendicular 






















SOLID GEOMETRY 


81 


XL. To describe the section of an ellipsoid when a 
section through the fixed axis, and the position of the 
line of the required section are given. 

Let A B C D (Fig. 51) be the section through the 
fixed axis of the ellipsoid, and F G the position of the 
line of the required section. Through the centre of 



the ellipsoid draw B D parallel to F G; bisect F G in 
H and draw A C perpendicular to F G; join B C, and 
from F draw F K, parallel to B C, and cutting A C pro¬ 
duced in K; and then will H K be the height of the 
semi-ellipse forming the section on F G. 















MODERN CARPENTRY 


Or, the section may be found by the method of or¬ 
dinates, thus: As the section of the ellipsoid on the 
line A C is a circle, from the point of intersection of 
B D and A C describe a semicircle A E C. Then on 
H G, the line of section, take any number of points, 1, 
m, p, and from them raise perpendiculars cutting the 
ellipse in q, r, s. From q, r, s draw lines perpendicular 
to A C, cutting it in the points 4, 5, 6; and again, from 
the intersection of B D and A C as centre draw, the 
arcs 41, 5 m, 6n, C o, cutting H G in 1, m, n, o; then 
II o, set off on the perpendicular from H to K, is the 
height of the section; and the heights IT n, M m, HI, 
set off on the perpendiculars from 1 to 3, n to 2 and 
p to 1, give the heights of the ordinates. 

XLI. To find the section of a cylindrical ring per¬ 
pendicular to the plane passing through the axis of the 
ring, the line of section being given. 

Let ABED (Fig. 52) be the section through the 
axis of the ring, A B a straight line passing through 
the concentric circles to the centre C, and D E be the 
line of section. On AB describe a semicircle; take in 
its circumference any points, as 1, 2, 3, 4, 5, &c., and 
draw the ordinates 1 f, 2 g, 3 h, 4 k, &c. Through the 
points f, g, h, k, 1, &c., where the ordinates meet the 
line A B, and from the centre C, draw concentric cir¬ 
cles, cutting the section line in m, n, o, p, q, &c. 
Through these points draw the lines m 1, n 2, o 3, &c., 
perpendicular to the section line, and transfer to them 
the heights of the ordinates of the semicircle f 1, g 2, 
&c.; then through the points 1, 2, 3, 4, &c., draw the 
curve D 5 E, which is the section required. 


SOLID GEOMETRY 


83 


Again, let R S be the line of the required section; 
then from the points t, u, v, w, c, x, d, &c., where the 
concentric circles cut this line, draw the lines t 1, u 2, 
v 3, &c., perpendicular to R S, and transfer to them the 
corresponding ordinates of the semicircle; and through 
the points 1. 2, 3, 4, e, 5, f, &c., draw the curve R e f S, 
which is the section required. 



XLII. To describe the section of a solid of resolu¬ 
tion the generating curve of which is an ogee. 

Let A D B (Fig. 53) be half the plan or base of the 
solid, A a b B the vertical section through its axis, and 
E F the line of section required. From G draw C 5 
perpendicular to E F, and bisecting it in m. In E m 






















84 


MODERN CARPENTRY 


take any number of points, g h k, &c., and through 
them draw the lines g 1, h 2, k 3, &c., perpendicular to 
E F. Then from C as a centre, through the points g, 
h, k, &c., draw concentric arcs cutting A B in r, s, t, u, 
y, and through these points draw the ordinates r 5, s 4, 
t 3, &c., perpendicular to A B. Transfer the heights 
of the ordinates on A B to the corresponding ordinates 
on each side of the centre of E F; and through the 
points 1, 2, 3, 4, 5, &c., draw the curve E 5 F, which 
is the section required. 



XLIII. To find the section of a solid of resolution, 
the generating curve of which is of a lancet form. 

Let A D B (Fig. 54) be the plan of half the base, A 
E B the vertical section, and F G the line of the re¬ 
quired section. The manner of finding the ordinates 














SOLID GEOMETRY 


85 


and transferring the heights is precisely the same as in 
the last problem. 

XLIV. To find the section of an ogee pyramid with 
a hexagonal base. 

Let ADEFB (Pig. 55) be the plan of the base of 
the pyramid, AabB a vertical section through its 
axis, and G LI the line of the required section. Draw 
the arrises C D, C E, C P. On the line of section G H, 
at the points of intersection of the arrises with it, and 



at some intermediate points k, m, o, q (the correspond¬ 
ing points k and q, and m and o, being equidistant 
from n), raise indefinite perpendiculars?. Through 
these points k, 1, m, n, o, p, q, draw lines parallel to 
the sides of the base, as shown by dotted lines; and 
from the points where these parallels meet the line 
A B, draw r 4, s 3, t 2, u 1, perpendicular to A B. These 
perpendiculars transferred to the ordinates n 4, m 3, 
o 5, 1 2, p 6, k 1, q 7, will give the points 1, 2, 3, 4, 5, 6, 
7, through which the section can be drawn. 














86 


MODERN CARPENTRY 


6. INTERSECTIONS OF CURVED SURFACES. 

When two solids having curved surfaces penetrate 
or intersect each otheB, the intersections of their sur¬ 
faces form curved lines of various kinds. Some of 
these, as the circle, the ellipse, &c., can be obtained in 
the plane; but the others cannot, and are named curves 
of double curvature. The solution of the following 
problems depends chiefly on the knowledge of how to 
obtain, in the most advantageous manner, the pro¬ 
jections of a point on a curved surface; and is in fact 
the application of the principles elucidated in the sev¬ 
eral previous problems. The manner of constructing 
the intersections of these curved surfaces which is the 
simplest and most general in its application, consists in 
conceiving the solids to which they belong as cut by 
planes according to certain conditions, more or less 
dependent on the nature of the surfaces. These sec¬ 
tion planes may be drawn parallel to one of the planes 
of the projection; and as all the points of intersection 
of the surfaces are found in the section planes, or on 
one of their projections, it is always easy to construct 
the curves by transferring these points to the other 
projection of the planes. 

XLV. The projection of two equal cylinders which 
intersect at right angles being given, to find the pro¬ 
jections of their intersections. 

Conceive, in the horizontal projection (Fig. 56), a 
series of vertical planes cutting, the cylinders parallel 
to their axis. The vertical projections of all the sec¬ 
tions will be so many right-angled parallelograms, sim- 


SOLID GEOMETRY 


87 


ilar to e f e f, which is the result of the section of the 
cylinder from surface to surface. The circumference 
of the second cylinder, whose axis is vertical, is cut by 
the same plane, which meets its upper surface at the 
two points g, h, and its under surface at two corre¬ 
sponding points. The vertical projections of these 
points are on the lines perpendicular to a b, raised on 



Fig. 56. 


each of them, so that upon the lines e f, e f, will be 
situated the intersections of these lines at the points g, 
h, and g, h and the same with other points i, k, 1, m. 
It is not necessary to draw a plan to find these pro¬ 
jections. All that is actually required is to draw the 
circle representing one of the bases (as n o) of the 
cylinder laid flat on the horizontal plane. Then to 
produce gh till it cuts the circle at the superior and 

































88 


MODERN CARPENTRY 


inferior points G, G, and to take the heights e G, e G, 
and carry them, upon a b, from g to g g and from h to 
h, h. 

Fig. 57 is the vertical projection made on the line 
xz. 



Fig. 57. 


XLVI. To construct the projections of two unequal 
cylinders whose axis intersect each other obliquely. 

Let A (Fig. 58) be the vertical projection of the two 
cylinders, and h S d e the horizontal projection of their 
axis. Conceive in the vertical projection, the cylin¬ 
ders cut by any number of horizontal planes; the hori¬ 
zontal projections of these planes will be rectangles, 
as in the previous example, and their sides will be 
parallel to the axis of the cylinders. 

The points of intersection of these lines will be the 
points sought. Without any previous operation, six of 
those points of intersection can be obtained. For ex¬ 
ample the point c is situated on d e, the highest point 
of the smaller cylinder; consequently, the horizontal 
projection of c is on d e, the horizontal projection of 
























SOLID GEOMETRY 


89 


d e, and it is also on the perpendicular let fall from c, 
that is to say, on the line c f parallel to the axis of the 
cylinder S h. The point sought will, therefore, be the 
intersection of those lines at c. In the same way i is 



obtained. The point j is on the line k 1, which is in the 
horizontal plane passing through the axis d e, the hori¬ 
zontal projections of kl are k 1, and its opposite mn; 
therefore in letting fall perpendiculars from j p, the 
intersections of these with k 1, m n, give the points j j, 


















90 


MODERN CARPENTRY 


p p. Thus six points are obtained. Take at pleasure 
an intermediate point q; through this point draw a 
line rs parallel to ab, which will be the vertical pro¬ 
jection of a horizontal plane cutting the cylinder in 
q. The horizontal projection of this section will be, as 
in the preceding examples, a rectangle which is ob¬ 
tained by taking, in the vertical projection, the height 
of the section plane above the axis d e, and carrying it 
on the base in the horizontal projection from G to T. 
Through T is then to be drawn the line Q U perpen¬ 
dicular to G T; and through Q and U the lines parallel 
to the axis; and the points in which these lines are in¬ 
tersected by the perpendiculars let fall from q u are 
the intermediate points required. Any number of inter¬ 
mediate points can thus be obtained; and the curve 
being drawn through them, the operation is completed. 

XLVII. To find the intersections of a sphere and a 
cylinder. 

Let e f c d and i k g h (Fig. 59) be the horizontal pro¬ 
jections of the sphere and cylinder respectively. Draw 
parallel to A B, as many vertical section planes are 
considered necessary, as e f, c d. These planes cut at 
the same time both the sphere and the cylinder, and 
the result of each section will be a circle in the case 
of the sphere, and a rectangle in the case of the cylin¬ 
der. Through each of the points of intersection g, h, 
i, k, and from the centre 1, draw indefinite lines per¬ 
pendicular to A B. Take the radius of the circles of 
the sphere proper to each of these sections and with 
them, from the centre 1, cut the correspondent perpen¬ 
diculars in gg, hh, ii, &c., and draw through these 
points the curves of intersection. 


SOLID GEOMETRY 


91 


XLVIII. To construct the intersection of two right 
cones with circular bases. 

The solution of this problem is founded on the 
knowledge of the means of obtaining on one of the 
projections of a cone a point given on the other. 



No. 1. Let AB (Fig. 60) be the common section of 
the two planes of projection, the circles g d e f and 
g h i k the horizontal projections of the given cones, 
and the triangles d i f and h 1 k their vertical projec¬ 
tions. Suppose these cones cut by a series of horizon¬ 
tal planes: each section will consist of two circles, the 



















92 


MODERN CARPENTRY 


intersections of which will be points of intersection 
of the conical surfaces. For example, the section made 
by a plane m n will have for its horizontal projections 
two circles of different diameters, the radius of the 
one being i m, and of the other 1 o. The intersecting 
points of these are p and q, and these points are com¬ 
mon to the two circumferences; and their vertical pro¬ 
jection on the plane m n will be p q. Thus, as many 
points may be found as is necessary to complete the 
curve. 

But there are certain points of intersection which 
cannot be rigorously established by this method with¬ 
out a great deal of manipulation. The point r in the 
figure is one of those; for it will be seen that at that 
point the two circles must be tangents to each other, 
and it would be difficult to fix the place of the sec¬ 
tion plane s t so exactly by trial, that it would just 
pass through the point. 

It will be seen that the point r must be situated in 
the horizontal projection of the line g i a perpendicu¬ 
lar i 1 equal to the height of the cone. From one raise 
a perpendicular and make it equal to the height of the 
second cone, and draw its side L i; and from the point 
of intersection R let fall a perpendicular on g i meet¬ 
ing it in r; through r draw an indefinite line perpen¬ 
dicular to A B, and set up on it from A B to r the 
height rR. The point r can also be obtained directly 
in the vertical projection by joining i g and 1 i as 
shown. 

Bisect r g in u, and from u as a centre, with the 
radius u r, describe a circle, the circumference of 
which will be the horizontal projection of the intersec- 


SOLID GEOMETRY 


93 

tion of the two cones. The vertical projection of this 
circle will be g p r q, and can be found by the method 
indicated above. 

No. 2. Conceive the horizontal projection (Fig. 61; 
No. .1) a vertical plane (j D cutting both cones through 
their axis: the sections will be two triangles, having 



the diameters of the bases of the cones as their bases, 
and the height of the cones as their height. And as in 
the example the cones are equal, the triangles will also 
be equal, as the triangles c e f, g f d, in the vertical 
projection. Conceive, now a number of inclined 
planes, as c n m, ck p, &c., passing through the differ¬ 
ent points of the base, but still passing through the 
summits of the cones; the sections which result will 
still be triangles (as has already been demonstrated), 
whose bases diminish in proportion as the planes re- 



























94 


MODERN CARPENTRY 


cede from the centres of the bases of the cones, until 
at length the plane becomes a tangent to both cones 
and the result is a tangent line whose projections are 
he, he, g e, f f. It will be observed that the circum¬ 
ferences of the bases cut each other at m and i, which 
are the first points of their intersections, whose veiti- 
cal projections are the point m merely. If the pro¬ 
jections^ the other points of intersection on the lines 
of the section planes are found (an operation present¬ 
ing no difficulty, and easily understood by the inspec¬ 
tion of the figure), it will be seen that the triangles 
n c m, m c o, kep, qer, &c., in the horizontal projec¬ 
tion, have for their vertical projections the triangles 
n e m, m o, k e p, &c., and that the intersections of the 
cones are in a plane perpendicular to both planes of 
projection, and the projections of the intersections are 
the right lines i m, m 3. From the known properties 
of the conic sections, the curve produced by this plane 
will be a hyperbola. Fig. 61, No. 2, gives tlie projec¬ 
tions of the cones on the line o x. 

No. 3. The next example (Fig. 62) differs from 
the last in the inequality of the size of the cones. Sup¬ 
pose an indefinite line C D to be the horizontal projec¬ 
tion of the vertical section plane, cutting the two cones 
through their axis e f. Conceive in this plane an in* 
definite line e f D, passing through the summits of the 
cones, the vertical projection of this line will be ef d: 
from d let fall on C D a perpendicular meeting it in D; 
this will be the point in which the line passing through 
the summits of the cones will meet the horizontal 
plane; -and it is through this point, and through the 
summits e and f, that the inclined section planes should 


SOLID GEOMETRY 


95 


be made to pass. The horizontal traces of these planes 
are 0 D, G D, &e.: 0 D is then; the trace of a tangent 
plane to the two conical surfaces 0 e, Pf; and the 
plane eGD cuts the greater cone, and forms by the 
section the triangles GeH in the horizontal, and geh 
in the vertical projection; and it cuts the lesser cone, 
and forms the triangles 1 f J, i f j. In the horizontal 



Fig. 62. 


projection it is seen that the sides H e, 1 f of the trian¬ 
gles intersect in k, which is therefore the horizontal 
projection of one of the points of intersection; and 
its vertical projection is k. In the same manner, other 
points can be found. It is seen at once that M, N, 1, 
are also points in the intersection. The curves traced 
through the points M k 1 N in the horizontal, and m k 1 
in the vertical projection, are the projections of the 
intersection of the two cones. 






96 


MODERN CARPENTRY 


7. COVERINGS OF SOLIDS. 

1. Regular Polyhedrons. A solid angle cannot be 
formed with fewer than three plane angles. The sim¬ 
plest solid is, therefore, the tetrahedron or pyramid 
having an equilateral triangle for its base, and its other 
three sides formed of similar triangles. 

The development of this figure (Fig. 63) is made by 
drawing the triangular base ABC, and then drawing 
around it the triangles forming the inclined sides. If 
the diagram is on flexible material, such as paper, then 
cut out, and the triangles folded on the lines A B, B C. 
C A, the solid figure will be constructed. 



Fig. 63. 


Fig. 64. 


The hexahedron, or cube, is composed of six equal 
squares (Fig. 64); the octahedron (Fig. 65) of eight 
equilateral triangles; the dodecahedron (Fig. 66) of 
twelve pentagons; the icosahedron (Fig. 67) of twenty 
















SOLID GEOMETRY 


97 


equilateral triangles. In these figures, A is the eleva¬ 
tion and B the development. 

The elements of these solids are the equilateral tri¬ 
angle, the square, and the pentagon. The irregular 
polyhedrons may be formed from those named, by cut¬ 
ting off the solid angles. Thus, in cutting off the 




angles of the tetrahedron, there results a polyhedron 
of eight faces, composed of four hexagons and four 
equilateral triangles. The cutting off the angles of the 
cube, in the same manner gives polyhedron of four- 











98 


MODERN CARPENTRY 


teen faces, composed of six octagons, united by eight 
equilateral triangles. 

The same operation performed on the octahedron 
produces fourteen faces, of which eight are hexagonal 
and six square; on the dodecahedron it gives thirty- 
two sides, namely, twelve decagons, and twenty trian¬ 
gles; on the icosahedron it gives thirty-two sides— 
twelve tentagons and twenty hexagons. This last ap¬ 
proaches almost to the globular form and can be rolled 
like a ball. 




The other solids which have plane surfaces are the 
pyramids and prisms. These may be regular and irreg¬ 
ular: they may have their axis perpendicular or in¬ 
clined ; they may be truncated or cut with a section, 
parallel or oblique, to their base. 





SOLID GEOMETRY 


99 


II. Pyramids. The development of a right pyra¬ 
mid, of which the base and the height are given, offers 
no difficulty. The operation consists (Fig. 68) in ele¬ 
vating on each side of the base, a triangle .having its 
height equal to the inclined height of each side, or, 
otherwise, connecting the sides, together as shown by 
the dotted lines. 



In an oblique pyramid the development is found as 
follows: Let abed (Fig. 69) be the plan of the base 
of the pyramid, a b c d its horizontal projection, and 
EFG its vertical projection. Then on the side dc 
construct the triangle c R d, making its height equal 
to the sloping side of the' pyramid ^ 2 . This triangle 
is the development of the side ^ P c of the pyramid. 










100 


MODERN CARPENTRY 


Then from d, with the radius E F, describe an arc 0; 
and from R, with the radius equal to the true length 
e G of the arris E G, describe another arc intersecting 
the last at 0. Join R 0, d 0; the triangle d R 0 will 
be the development of the side a P d. In the same way, 
describe the triangle c R T, for the development of the 
side b P c. From R, again, with the same radius R 0, 
describe an arc S, which intersect by an arc described 
from 0 with the radius a b; and the triangle 0 R S 
will be the development of the side a P b. 

If the pyramid is truncated by a plane w a parallel 
to the base, the development of that line is obtained by 
setting off from R on R c, and R d, the true length of 
the arris G a in x and 2, and on R S, R 0, and R T, the 
true length of the arris G w in 4, 3, 1; and drawing the 
lines 1 x, x2, 2 3, 3 4, parallel to the base of the respect¬ 
ive triangles T R c, c R d, d R 0, 0 R S. If it is trun¬ 
cated by a plane wy, perpendicular to the axis, then 
from the point R, with the radius equal to the true 
length of the arris G w, or G y, describe an arc 1 4, and 
inscribe in it the sides of the polygon forming the 
pyramid. 

III. Prisms. In a right prism, the faces being all 
perpendicular to the bases which truncate the solid, it 
results that their development is a rectangle composed 
of all the faces joined together, and bounded by two 
parallel lines equal in length to the contour of the 
bases. Thus, in Fig. 70, a b c d is the base, and b e the 
height of the prism; the four sides will form the rec¬ 
tangle b e f g, and e h i k will be the top of the prism. 
The full lines show another method of development. 

When a prism is inclined, the faces form different 


SOLID GEOMETRY 


101 


angles with the lines of the contours of the bases; 
whence there results a development, the extremities 
of which are bounded by lines forming parts of poly¬ 
gons. 



Fig. 70. 


After having drawn the line C C (Fig. 71), which in¬ 
dicates the axis of the prism and the lines A B, D E, 
the surfaces which terminate it, describe on the middle 
of the axis the polygon forming the plan of the prism, 
taken perpendicularly to the axis, and indicated by the 
figures 1 and 8. Produce the sides 1 2, 6 5, parallel to 
the axis, until they meet the lines A B, D E. These 
lines then indicate the four arrises of the prism, cor¬ 
responding to the angles 1 2 5 6. Through the points 

















102 


MODERN CARPENTRY 


8 3 7 4 draw lines parallel to the axis meeting A B> 
D E in F H, G L. These lines represent the four 
arrises 8 3 7 4.* 

In this profile the sides of the plan of the polygon 
12345678 give the width of the faces of the prism, 



and the lines AD, FH, GL, BE their length. From 
this profile can be drawn the horizontal projection, in 
the manner shown below. To trace the development of 
this prism on a sheet of paper, so that it can be folded 

♦In Fig. 75 another example is given, but as the method of 
procedure is the same as in Fig. 71, detailed description is un¬ 
necessary. 


















SOLID GEOMETRY 


103 


together to form the solid, proceed thus: On the 
middle of C C raise an indefinite perpendicular M N. 
On that line set off the width of the faces of the prism, 
indicated by the polygon, in the points 01234567 8.' 
Through these points draw lines parallel to the axis, 
and upon them set off the lengths of the lines in profile, 
thus: From the points O, 1, and 8, set off the length 
M D in the points D D D; from 2 and 7, set off a H in 
H; from 3 and 6, set off b L in L and L; and so on. 
Draw the lines DD, DHLE, EE, EL HD, for the 
contour of the upper part of the prism. To obtain the 
contour of the lower portion, set off the length MA 
from 0, 1, and 8 to A A A, the length a F from 2 and 
7 to F and F, the length b G from 3 and 6 to G and G, 
and so on; and draw A A, A F G B, B B, BGFA, to 
complete the contour. The development is completed 
by making on B B and E E the polygons 12 3 4 5 6 BB, 
12 3 4 5 6 EE, similar to the polygons of the horizon¬ 
tal projection. 

IV. Cylinders. Cylinders may be considered as 
prisms, of which the base is composed of an infinite 
number of sides. Thus we shall obtain graphically the 
development of a right cylinder by a rectangle of the 
same height, and of a length equal to the circumference 
of the circle, which serves as its base. 


To find the covering of a right cylinder. 

Let A B C D (Fig. 72) be the seat or generating sec¬ 
tion. On A D describe the semicircle A 5 D, represent¬ 
ing the vertical section of half the cylinder, and divide 
its circumference into any number of parts, 1, 2, 3, 4, 


104 


MODERN CARPENTRY 


5, &c., and transfer those divisions to the lines A D 
and B C produced; then the parallelogram D C G P will 
be the covering of one half the cylinder. 



To find the edge of the covering when it is oblique 
in regard to the sides of the cylinder. 

Let AB CD (Fig. 73) be the seat of the generating 
section, the edge B C being oblique to the sides A B, 
D C. Draw the semicircle A 5 D, and divide it into any 



number of parts as before; and through the divisions 
draw lines at right angles to AD, producing them to 
meet B C in r s, t, u, v, &c. produce A D, and the lines 
la, 2 b, 3 c, &c., perpendicular to' D F. To these lines 













































SOLID GEOMETRY 


105 


transfer the length of the corresponding lines inter¬ 
cepted between A D and B C, that is, to la transfer the 
length p z, to 2 b transfer o y, and so on, by drawing 
the lines z a, y b, x c, &c., parallel to A F. Through the 
points thus obtained, draw the curved line C a b c, &c., 
to G; then shall DFCG be the development of the 
covering of the semi-cylinder A B C D. 



To find the covering of a cylinder contained between 
two oblique parallel planes. 

Let AB CD (Fig. 74) be the seat of the generating 
section. From A draw A G perpendicular to A B, and 
produce C D to meet it in E. On A E describe the semi¬ 
circle, and transfer its perimeter to E G, by dividing 
it into equal parts, and setting off corresponding di¬ 
visions on E G. Through the divisions of the semicircle 
draw lines at right angles to AE, producing them to 

































MODERN CARPENTRY 


meet the lines A D and B C, in i, k, 1, m, &c. Through 
the divisions on E G draw lines perpendicular to it; 
then through the intersections of the ordinates of the 
semicircle, with the line A D, draw the lines i a, k z, 1 y, 





Fig. 75. 


&c., parallel to A G and where these intersect the per¬ 
pendiculars from E G, in the points a, z, y, x, w, u, &c., 
trace a curved line GD, and draw parallel to it the 
curved line H C; then will D C H G be the development 
of the covering of the semi-cylinder A B C D. 

To find the covering of a semi-cylindric surface 
bounded by two curved lines. 

The construction to obtain the developments of these 
coverings (Figs. 76 and 77) is precisely similar to that 
described in Fig. 74. 







SOLID GEOMETRY 


107 




Fig. 77. 























































































108 


MODERN CARPENTRY 


V. Cones. We have considered cylinders as prisms 
with polygonal bases. In a similar manner we may re¬ 
gard cones as pyramids. 

In right pyramids, with regular symmetrical bases,. 


O' 



Fig. 78. 


as the lines of the arrises extending from the summit 
to the base are equal, and as the sides of the polygons 
forming the base are also equal, their developed sur¬ 
faces will be composed of similar and equal isosceles 
triangles, which, as we have seen (Fig. 78, a, b, c, d), 
will, when united, form a part of a regular polygon 



































SOLID GEOMETRY 


109 


inscribed in a circle, of which the inclined sides of the 
polygon form the radii. Thus in considering the base 
of the cone KH (Fig. 78) as a regular polygon of an 
infinite number of sides, its development will be found 
in the sector of a circle, MAFBM (No. 3), of which 
the radius equals the side of the cone K G (No. 1), and 
the arc equals the circumference of the circle forming 
its base (No. 2). 

To trace on the development of the covering, the 
curves of the ellipse, parabola, and hyperbola, which 
are the result of the sections of the cone by the lines 
D I, E F, I G, it is necessary to divide the circumfer¬ 
ence of the base AFBM (No. 2) into equal parts, as 
1, 2, 3, &c., and from these to draw radii to the centre 
C, which is the horizontal projection of the vertex of 
the cone; then to carry these divisions to the common 
intersection line KII, and from their terminations 
there to draw lines to the vertex G, in the vertical pro¬ 
jection No. 1. These lines cut the intersecting planes, 
forming the ellipse, parabola, and hyperbola, and by 
the aid of the intersections we obtain the horizontal 
projection of these figures in No. 2—the parabola 
passing through M E F, the hyperbola through GIL, 
and the ellipse through D I. 

To obtain points in the circumference of the ellipse 
upon the development, through the points of inter¬ 
section o, p, q, r, &c., draw lines parallel to K H, car¬ 
rying the heights to the side of the cone G H, in the 
points 1, 2, 3, 4, 5, 6, 7, and transfer the lengths G 1, 
G 2, G 3, &c. to G 1, G 2, G 3, G 4, &c., on the radii of 
the development in No. 3 and through the points thus 
obtained draw the curve z D 1 X. 


110 


MODERN CARPENTRY 


To obtain the parabola and hyperbola, proceed in 
the same manner, by drawing parallels to the base 
KH, through the points of intersection; and transfer¬ 
ring the lengths thus obtained on the sides of the cone 
G K, G H, to the radii in the development. 

Nos. 4, and 5 give the vertical projections of the hy¬ 
perbola and parabola respectively. 



To find the covering of the frustum of a cone, the 
section being made by a plane perpendicular to the 
axis. 

Let ACEF (Fig. 79) be the generating section of 














SOLID GEOMETRY 


111 


the frustum. On A C describe the semicircle ABC, 
and produce the sides AE and C F to D. From the 
centre D, with the radius D C, describe the arc C IT; 
and from the same centre with the radius D F, describe 


D 



the arc F G. Divide the semicircle ABC into any 
number of equal parts, and run the same divisions 
along the arc C H; draw fhe line H D, cutting E G in 
G; then shall C H G F be half the development of the 
covering of the frustum A C F E. 

















112 


MODERN CARPENTRY 


To find the covering of the frustum of a cone, the 
section being made by a plane not perpendicular to 
the axis. 

Let ACFE (Fig. 80) be the frustum. Proceed as 
in the last problem to find the development of the cov¬ 
ering of the semi-cone. Then—to determine the edge 
of the covering of the line EF—from the points P, q, 
r, s, t, &c., draw lines perpendicular to E F, cutting 
AC in y, x, w, v, u; and the length u t transferred 
from 1 to a, v s, transferred from 2 to b, and so on, will 
give a, b, c, d, e, &c., points on the edge of the cover¬ 
ing. 

To find the covering of the frustum of a cone, when 
cut by two cylindrical surfaces perpendicular to the 
generating section, 

Let AEFC (Fig. 81) be the given frustum, and 
AkC, EpF, the given cylindrical surfaces. Produce 
A E, C F, till they meet in the point D. Describe the 
semicircle ABC, and divide it into any number of 
equal parts, and transfer the divisions to the arc C H, 
described from D, with the radius D C. Through the 
divisions in the semicircle 1, 2, 3, 4, &c., draw lines per¬ 
pendicular to A C, and through the points where they 
intersect A C draw lines to the summit D. Draw lines 
also through the points 1, 2, 3, 4, 5, &c., of the arc C H, 
to the summit D; then through the intersections of the 
lines from A C to D, with the seats of the cylindrical 
surfaces k, 1, m, n, o, and p, q,r, s, t, draw lines parallel 
to A C, cutting C D; and from the points of intersec¬ 
tion in C D and from the centre D, describe arcs cut¬ 
ting the radial lines in the sector DCHin u, v, w, x, y, 


SOLID GEOMETRY 


113 


&c., and a, b, c, d, e, &c., and curves traced through 
the intersections will give the form of the covering. 

VI. Spheres, Ellipsoids, &c. The development of 
the sphere, and of other surfaces of double curvature, 



Fig. 81. 


is impossible, except on the supposition of their being 
composed of a great number of small faces, either 
plane, or of a simple curvature, as the cylinder and the 
















114 


MODERN CARPENTRY 



cone. Thus, the sphere or spheroid may be considered 
as a polyhedron, terminated, 1st, by a great number 
of plane faces, formed by truncated pyramids, of which 
the base is a polygon, as in Fig. 82; 2nd, by parts of 
















SOLID GEOMETRY 


.115 


truncated cones forming zones, as in Fig. 83, the part 
above AB being the vertical projection, and the part 
below AB the horizontal projection; 3rd, by parts of 
cylinders cut in gores, forming flat sides, which dimin¬ 
ish in width, as in Fig. 84. 



In reducing the spheres, or spheroid, to a polyhedron 
with flat sides, two methods may be adopted, which dif¬ 
fer only in the manner of arranging the developed 
faces. 

The most simple method is by parallel circles, and 
others perpendicular to them, which cut them in two 
opposite points, as in the lines on a terrestrial globe. If 
we suppose that these divisions, in place of being 
circles, are polygons of the same number of sides, there 
will result a polyhedron, like that represented in Fig. 
82, of which the half, A D B, shows the geometrical ele¬ 
vation, and the other half, A E B, the plan. 

To find the development, first obtain the summits 










116 


MODERN CARPENTRY 


P, q, r, s, of the truncated pyramids, which from the 
demi-polyhedron A D B, by producing the sides A 1, 
12, 23, 34, until they meet the axis E D produced; 
then from the points p, q, r, s, and with the radii P A, 
PI, q I q 2, r 2, r 3, and s 3, s 4, describe the indefinite 
arcs A B, 1 b, 1 b, 2 f, 2 f, e q, 3 g, 4 h, and from D de¬ 
scribe the arc 4 h; upon all these arcs set off the divi¬ 
sions of the demi-polygons A E B, and draw tfie lines to 
the summits p, q, r, s, and D, from all the points so set 
out, as A, 1, 2, 3, 4, &c., from each truncated pyramid. 
These lines will represent for every band or zona the 
faces of the truncated pyramids of which they consti¬ 
tute a part. 

The development can also be made by drawing 
through the centre of each side of the polygon A E B, 
indefinite perpendiculars, and setting out upon them 
the heights of the faces in the elevation, A1 2 3 4 D, 
and through the points thus obtained drawing paral¬ 
lels to the base. On each of these parallels then set 
out the widths, h, i, k, 1, d, of the corresponding faces 
(e, e, e, &c.) in the plan, and there will be thus formed 
trapeziums and triangles, as in the first development, 
but arranged differently. This method is used in con¬ 
structing geographical globes, the other is more con¬ 
venient in finding the stones of a spherical dome. 

The development of the sphere by reducing it to 
conical zones (Fig. 83) is accomplished in the same 
manner as the reduction to truncated pyramids, with 
this difference, that the developments of the arrises, 
indicated byA12345Bin Fig. 82, are arcs of circles 
described from the summits of cones, in place of being 
polygons. 


SOLID GEOMETRY 


117 


The development of the sphere reduced into parts 
of cylinders, cut in gores (Fig. 84), is produced by the 
second method described, but in place of joining, by 
straight lines, the points E, h, i, k, 1, d (Fig. 82), we 



Fig. 85. 

unite them by curves. This last method is used in 
tracing the development of caissons in spherical or 
spheroidal vaults. 

To find the covering of a segmental dome. 

In Fig. 85, No. 1 is the plan, and No. 2 the elevation 
of a segmental dome. Through the centre of the plan 















118 


MODERN CARPENTRY 


E draw the diameter A C, and the diameter B D per¬ 
pendicular to A C, and produce B D to I. Let D E rep¬ 
resent the base of semi-section of the dome; upon D E 
describe the arc D k with the same radius as the 
arc F G H (No. 2); divide the arc D k into any num¬ 
ber of equal parts, 1, 2, 3, 4, 5, and extend the division 
upon the right line D I, making the right line D I equal 
in length, and similar in its divisions, to the arc D k: 
from the points of division, 1, 2, 3, 4, 5, in the arc D k, 
draw lines perpendicular to D E, cutting it in the 
points q, r, s, &c. Upon the circumference of the plan 
No. 1, set off the breadth of the gores or boards 1 m, 
m n, n o, o p, &c.; and from the points 1, m, n, o, p, 
draw right lines through the centre E: from E describe 
concentric arcs q v, r u, st, &c., and from 1 describe 
concentric arcs through the points D, 1, 2, 3, 4, 5,: 
1 m, being the given breadth of the base, make 1 w 
equal to q v, 2x equal to r u, 3 y equal to st, &c.; draw 
the curved line through the points 1, w, x, y, &c., to 1, 
which will give one edge of the board or gore to coin¬ 
cide with the line 1 E. The other edge being similar, 
it will be found by making the distances from the 
centre line D 1 respectively equal. The seats of the 
different boards or gores on the elevation are found 
by the perpendicular dotted lines, pp, oo, nn, m m, &c. 


To find the covering of a semicircular dome. 

Fig. 86, Nos. 1 and 2.—The procedure here is more 
simple than in the case of the segmental dome, as, the 
horizontal and vertical sections being alike, the or¬ 
dinates are obtained at once. 


SOLID GEOMETRY 


119 


To find the covering of a semicircular dome when it 
is required to cover the dome horizontally. 


G 



Let ABC (Fig. 87) be a vertical section through the 
axis of a circular dome, and let it be required to cover 
this dome horizontally. Bisect the base in the point 


















120 


MODERN CARPENTRY 


D, and draw D B E perpendicular to A C, cutting the 
circumference in B. Now divide the arc B C into 
equal parts, so that each part will be rather less than 



Fig. 87. 

the width of the a board; and join the points division 
by straight lines, which will form an inscribed polygon 
of so many sides; and through these points draw line 
parallel to the base A C, meeting the opposite sides of 























SOLID GEOMETRY 


121 


the circumference. The trapezoids formed by the sides 
of the polygon and the horizontal lines may then be 
regarded as the sections of so many frustums of cones; 
whence results the following mode of procedure, in 



accordance with the introductory illustration Fig. 82 : 
*—Produce, until they meet the line D E, the lines g f, 
f, n, &c., forming the sides of the polygon. Then, to de¬ 
scribe a board which corresponds to the surface’ of one 
of the zones, as f g, of which the trapezoid m 1 f g is a 
















122 


MODERN CARPENTRY 


section,—from the point h, where the line f g produced 
meets D E, with the radii h f, hg, describe two arcs, 
and cut off the end of the board k on the line of a 
radius h k. 

To obtain the true length of the board, proceed as in 
Fig. 89. The other boards are described in the same 
manner. 



To find the covering of an elliptic dome. 

Let A B C D (Fig. 90) be the plan, and FG H the 
elevation of the dome. Divide the elliptical quadrant 
F G (No. 2) into any number of equal parts in 1, 2, 3, 
4, 5, and draw through the points of division lines 











SOLID GEOMETRY 


123 


perpendicular to F H, and produced to A C (No. 1), 
meeting it in i, k, m, n,: these divisions are transferred 
by the dotted arcs to the gore b E c and the remainder 
of the process is as in Figs. 85 and 86. 

To find the covering boards of an ellipsoidal dome. 

Let ABCD (No. 1, Fig. 89) be the plan of the dome, 
and FGH (No. 2) the vertical section through its 
major axis. Produce F H indefinitely to n; divide the 
circumference, as before, into any number of equal 
parts, and join the divisions by straight lines, as p m, 
ml, &c. Then, describe on a board, produce the line 
forming one of the sides of the polygon, such as 1 m, to 
meet Fninn; and from n, with the radii n m, n 1 
describe two arcs forming the sides of the board, and 
cut off the board on the line of the radius n o. Lines 
drawn through the points of divisions at right angles 
to the axis, until they meet the circumference ADC 
of the plan, will give the plan of the boarding. 

To find the covering of an ellipsoidal dome in gores. 

Let the ellipse ABCD (Fig. 90, No. 1) be the plan 
of the dome, A C its major axis and B D its minor axis; 
and let A B C (No. 2) be its elevation. Then, first, to 
describe on the plan and elevation the lines of the 
gores, proceed thusThrough the line AC (No. 1) 
produced at H, draw the line E G perpendicular to it, 
and draw BE, D G, parallel to the axis AC, cutting 
EG, then will EG be the length of the axis minor, 
on which is to be described the semi-circle E F G, rep¬ 
resenting a section of the dome on a vertical plane 
passing through the axis minor. 

Divide the circumference of the semi-circle into any 
number of equal parts, representing the widths of the 


124 


MODERN CARPENTRY 


covering boards on the line BD; and through the 
points of division 1, 2, 3, 4, 5, draw lines parallel to the 
axis A C, cutting the line B D in 1, 2, 3, 4, 5. Divide 
the quadrant of the ellipse C D (No. 1) into any num¬ 
ber of equal parts in e, f, g, h, and through these points 
draw the lines e a, f b, gc, h d in both diagrams, per- 





Fig. 90. 


pendicular to A C, and these lines will then be the seats 
of vertical sections through the dome, parallel to 
EFG. Through the points e, f, g, h (No. 1) draw 
lines parallel to the axis A C, cutting E G in o, n, m, k ; 






































































SOLID GEOMETRY 


125 


and from H, with the radii Ho, Hn, Hm, Hk, de¬ 
scribe concentric circles o 9 9 p, n 8 8q, m z y r, &c. To 
find the diminished width of each gore at the sections 
a, e, bf, eg, dh:—Through the divisions of the semi¬ 
circle, 1, 2, 3, 4, 5, draw the radii Is, 2t, 3u, 4v, 5w, 6x; 
then by drawing through the intersections of these 
radii with the concentric circles, lines parallel to H F, 
to meet the section lines corresponding to the circles, 
the width of the gores at each section will be obtained; 
and curves through these points will give the repre¬ 
sentation of the lines of the gores of the plan. 

In No. 2 the intersections of the lines are more 
clearly shown. The quadrant EGF is half the end 
elevation of the dome, and is divided as in No. 1. The 
parallel lines 5 5, 4 4, 3 3, show how the divisions of 
the arc of the quadrant are transferred to the line D B, 
and the other parallels ah, b k, cl, dm, are drawn 
from the divisions in the circumference of the ellipse 
to the line EG, and give the radii of the arcs m, 1 o, 
k p, h q. 

To describe one of the gores draw any line AB 
(No. 3), and make it equal in length to the circumfer¬ 
ence of the semi-ellipse ADC, by setting out on it the 
divisions 1, 2, 3, 4, 5, &c., corresponding to the divisions 
C h, h g, g f, &c., of the ellipse: draw through those 
divisions lines perpendicular to AB. Then from the 
semi-circle (No. 1) transfer to these perpendiculars the 
widths 6 5 to g n, 9 9 to f m, 8 8 to e 1, y z to d k, and 
x w to c h, and join Ac, d d, de, e f, f g, A h, h k, k 1, 

1 m, and m n,; which will give the boundary lines of 
one-half of the gore, and the other half is obtained in 
the same manner. 


126 


MODERN CARPENTRY 


To describe the covering of an ellipsoidal dome with 
horizontal boards of equal width. 

Let A B C D (No. 1, Fig. 91) be the plan of the dome, 
ABC (No. 2) the section on its major axis, and L M N 
the section on its minor axis. Draw the circumscribing 
parallelogram of the ellipse, namely, FGHK (No. 1), 
and its diagonals FHGK. In No. 2 divide the cir¬ 
cumference into equal parts, 1, 2, 3, 4, representing the 
number of covering boards, and through the points of 
division draw lines 18, 2 7, &c., parallel to A C. 
Through the points of divisions draw 1 p, 2 t, 3 x, &c., 
perpendicular to A C, cutting the diagonals of the cir¬ 
cumscribing parallelogram of the ellipse (No. 1), 
and meeting its major axis in p, t, x, &c. Complete 
the parallelograms, and inscribe ellipses therein cor¬ 
responding to the lines of the covering. Produce the 
sides of the parallelograms to intersect the circumfer¬ 
ence of the section on the minor axis of the ellipse in 
1, 2, 3, 4, and lines drawn through these parallel to 
L N, will give the representation of the covering boards 
in that section. To find the development of the cover¬ 
ing, produce the axis D B, in No. 2, indefinitely. Join 
by a straight line the divisions 1 2 in the circumfer¬ 
ence, and produce the line indefinitely, making e k 
equal to e 2, and k g equal to 12; 1 2e k g will be the 
axis major of the ellipses of the covering 12 7 8. Join 
also the corresponding divisions in the circumference 
of the section on the minor axis, and produce the line 
12b to meet the axis produced; and the length of 
this line will be the semi-axis minor, e h, of the ellipse, 
2 h k, while the width f h will be equal to the division 
12 in N M L. 


SOLID GEOMETRY 


127 


























































128 


MODERN CARPENTRY 


To find the covering 1 of an annular vault. 

Let ACKGEFA (Pig. 92) be the generating sec¬ 
tion of the vault. On AC describe a semi-circle ABC, 
and divide its circumference into equal parts, repre¬ 
senting the boards of the covering. From the divisions 
of the semi-circle, b, m, t, &c., from the centre D of 



the annulus, with the radii Dr, D s, &c., describe the 
concentric circles, s q, &c., representing the covering 
boards in plan. Through the centre D draw H K per¬ 
pendicular to G C, indefinitely extending it through K. 
Join the points of division of the semi-circle, Ab, b m, 
m t, by straight lines, and produce them until they cut 



















SOLID GEOMETRY 


129 


the line K H as m b n, t m u, when the points n, u, &c., 
are the centres from which the curves of the covering 
boards mo, tv, &c., are described. 



M 

Fig. 93. 


To find the covering of an ogee dome, hexagonal in 
plan. 

Let A B C D E F (No. 1, Fig. 93) be the plan of the 
dome, and H K'L (No. 2) the elevation, on the di¬ 
ameter F C. Divide H K into any number of equal 
parts in 1, 2, 3, 4, 5, k; and through these draw per¬ 
pendiculars to HL, and produce them to meet FC 

























130 


MODERN CARPENTRY 


(No. 1) in 1, mm, n, o, p, G. Through these points 
draw lines 1 d, me, n f, &c., parallel to the side F E of 
the hexagon : bisect the side FE in N, and draw G N, 
which will be the* seat of a section of the dome, at right 
angles to the side E F. To find this section nothing 
more is required than to set up on N G from the points 
t, u, v, &c., the .heights of the corresponding ordinates 
ql, r2, s3, &c., of the elevation (No. 2) to draw the 
ogee curve N1 2 3 4 5 p, and then to use the divisions 
in this curve to form the gore or covering of one side 
EghkMD. 


PART II. 

PRACTICAL SOLUTIONS. 

Having taken a thorough course in Solid Geometry, 
the student should be now prepared to solve almost 
any problem in practical construction, almost as soon 
as they present themselves. The various problems in 
construction, however, are so numerous, and in many 
cases, so intricate, that the student will often be con¬ 
fronted by problems which will require so many appli¬ 
cations of the rules he has been taught, in different 
forms, that without some helping guidance, he will fail 
to see exactly what to do. 

The following examples, with their explanations, are 
intended to give him the aid necessary to solve many 
difficult problems, and equip him with the means of still 
further investigation and sure results. 

It is often necessary for the workman to find the 
exact stretchout or length of a straight line that shall 
equal the quadrant or a semi-circle. 

To accomplish this:—Take AB radius, and A cen¬ 
tre; intersect the circle at C; join it and B; draw 
from D, parallel with C B, cutting at H; then A H will 
be found equal to curve A D. Fig. 1. 

This method is sojnewhat different to that already 
given; both, however, are practical. 

Fig. 2. To find a straight line which is equal to the 
circumference of a circle. Draw from the centre, O, 
131 


132 MODERN CARPENTRY 


D 




Fig. 2. 







PRACTICAL SOLUTIONS 


133 


any right angle, cutting at J and V; join JV; draw 
from 0 parallel with JV; square down from J, cut¬ 
ting at N; joint it and V; then four times N V will be 
found to equal the circumference. 

How to find the mitres for intersecting straight and 
circular mouldings. 

Figure. 3 shows the form of an irregular piece of 
framing or other work, which requires to have mould¬ 
ings mitre and properly intersect. 

The usual way of doing this is to bisect each angle, 
or to lay two pieces of moulding against the sides of 
framing, and mark along the edge of each piece, thus 
making an intersectic or point, so that by drawing 
through it to the next, point, which is the angle of 
framing, the direction of mitre is obtained. This pro¬ 
cess, however, is not the quickest and best by any 
means. The most simple and correct method is to ex¬ 
tend the sides A L and P H. 

Now suppose we wish to find a mitre from L; take 
it as centre, and with any radius, as K, draw the circle, 
cutting at J; join it and K draw from L parallel with 
J K, and we have the mitre at once. 

Now come to angle on the right; here take II as 
centre, and with any radius, as E, draw the circle, 
cutting at F; join it and E; draw from H parallel with 
E F, and you will find a correct mitre. 

The next question is the intersectioa of straight and 
circular mouldings. 

In the present case an extreme curve is given, in 
order to show the direction of mitre here, which is 
simply on the principle of finding a centre, for three 
points not on a straight line. For example, A B C are 


134 MODERN CARPENTRY 



Pig. 









PRACTICAL SOLUTIONS 135 

points; bisect A B and B C; drawn through intersec¬ 
tions thus made, and lines meeting in point D give a 
centre, from which strike the circular mitre as shown. 

Here it may be stated that in some cases a straight 
line for mitres will answer; this means when the curve 
is a quadrant or less. 

Fig. 4 shows the intersections of rake and level 
mouldings for pediments. 

The moulding on the rake, increases in width, and 
is entirely different from that on the level, yet both 
mitre, and intersect, the rake moulding being worked 
to suit the level. If the curves of Fig. 4 are struck from 
centres as shown, then by the same rule, the rake 
moulding is also struck from centres. 

Take any point in the curve, as C; square up from it, 
cutting at B; draw from C parallel to SL; join L K, 
which bisect at N; make E D equal to A B on the right; 
join L D and D N; bisect L D, also D N; draw through 
intersections thus made, and the lines meeting in F, 
give a point from which draw through N; make N J 
equal H F; then F and J are centres, from which strike 
the curve, and it will be found to exactly intersect 
with that of Fig. 4. 

Both mouldings here are shown as solid, and of the 
same thickness. This is done for the purpose of mak¬ 
ing the drawing more plain and easily understood; 
but bear in mind that all crown mouldings are gener¬ 
ally sprung. 

To find the form of a sprung* or solid moulding on any 
rake without the use of either ordinates or centres. 

It may not be generally known, that if a level mould¬ 
ing is cut to a mitre, that the extreme parts of mitre, 


MODERN CARPENTRY 




Fig. 












PRACTICAL SOLUTIONS 137 

when in a certain position, will instantly give the 
exact form of a rake moulding, and it will intersect, 
and mitre correctly with that of level moulding. To 
do this, take the level piece which has been mitred; 
lay its flat surface on the drawing; make its point P 
at Pig. 5, stand opposite point P at Fig. 6; keep the 
outer edge fair with line N L. The piece being in this 
position, take a marker, hold it plumb against the 
mitre, and in this way, prick off any number of points 
as shown, through which trace the curve line, and the 
result is a correct pattern by which the rake moulding 
is worked. 

A moment’s consideration will convince us that this 
simple method must give the exact form of any rake 
moulding to intersect with one on the level. 

To cut the mitres and dispense with the use of a 
box, this method will be found quick and off-hand. 
Take, for example, the back level moulding, and square 
over on its top edge any line, as that of FN; con¬ 
tinue it across the back to TI; make IIV equal T L 
above, and from V, square over lower edge II K - Now 
take bevel 2 froip above, and apply it on top edge, 
as shown; mark FL; then join L V; cut through these 
lines from the back,, and the mitre is complete. 

To cut the mitre on the rake moulding, square over 
any line on its back, as that of H J; continue it across 
the top and lower edge; take bevel X, shown above 
Fig. 5, apply it here on top edge, and mark DA; take 
the same bevel, and apply it on small square at E, and 
mark E 2. 

We now want the plumb cut on lower edge J K, and 
the same cut on front edge N P, shown at Fig. 6. Take 


138 MODERN CARPENTRY 



Fig. 5. Fig. 6. 

































PRACTICAL SOLUTIONS 


139 


bevel W above Fig. 5, apply it here and mark 2 B; join 
B A; this done, apply the same bevel on front edge 
NP, and mark the plumb cut, it being parallel with 
that of 2B here, or K J, Fig. 5; now cut through lines 
on the back, and the mitre is complete. 

It has already been shown, that we dispense with 
making or using a box for mitreing sprung mouldings. 

In this case, the front edge or upper member, stands 
parallel with face of wall, so that bevel X being ap- 
piled, gives the plumb cut; then the cut on top edge 
is square with face of wall. This shows, that we have 
only to find the direction of a cut on the back of mould¬ 
ing to make the mitre. 

To do this, take any point as R; draw from it square 
with rake of gable. Now mark sections of moulding, 
as shown, its back parallel with RF; draw from D 
square with E N; extend the rake to cut line from D at 
K; this done, take any point on the rake, say L; draw 
from it parallel with R F, cutting at K; take it as 
centre and L as radius, and draw the arc of a circle; 
with same radius return to K on the right; take it as 
centre, and draw the arc L H; make the first arc equal 
it; then draw from H parallel with L C, cutting at 
C J; draw from it square with rake, cutting at C, and 
join C K. This gives bevel W for cut on back of mould¬ 
ing. 

A most perfect illustration of this may be had by 
having the drawing on card-board, and cutting it clear 
through all the outer lines, including that of the mould¬ 
ing on lines F D N E, making a hinge by a slight cut 
on line R F; also make 'a hinge of line R A, by a slight 
cut on the back, and in like manner make front edge 


140 


MODERN CARPENTRY 


work on a hinge by a slight cut on line F Y. This cut 
is made on top surface. Perform the same operation 



on the left. All the cuts being made, raise both sides 
on hinges AR and A 2; push the sections of mould¬ 
ings on right and left from you; make front edge rest 









PRACTICAL SOLUTIONS 


141 


on F D. Now bring mitres together, and we have a 
practical illustration of mitreing sprung mouldings on 
the rake. (See Fig. 7.) 


PROJECTION OF SOLIDS. 

The following illustrations will be found by the stu¬ 
dent almost indispensable in the construction of vari¬ 
ous objects, and they will open the door to many more. 


j> c x 



Fig. 8. 


Fig. 8 shows the projection of a solid. This means 
the section of anything that is cut by a plane not paral¬ 
lel to its base; or, to put this in a more practical way 
—take a square bar of wood and cut it in the direction 
of BE; the section it makes is shown by BEFH; 
simple as this is, it still gi'ves the idea of what is mean> 
by projection. 






142 


MODERN CARPENTRY 


Fig. 9 shows the section of a square bar which has 
been cut by two unequal pitches, say in the direction 
of bevels J and H; the line C B is called the seat; 
from it all measurements are taken and transferred to 
lines that are square with the pitch A B; this pitch 
may be called a diameter, because it and the ordinate 
A E are at right angles. 


JS 



Fig. 10 shows the sides of a square bar which may 
be any length. The bar is to stand perpendicular, and 
pass through a plank that inclines at A B; the learner 







PRACTICAL SOLUTIONS 143 

is now required to show on the surface of plank the 
shape of a mortise that shall exactly fit the bar. 



Fig. 10. 


To solve this the student is left to exercise his own 
intelligence. 

The problem may be clearly demonstrated by a 
card-board model; it being cut, and the parts pro- 






144 


MODERN CARPENTRY 


jected from the flat surface will represent the plank, 
and show the mortise in it standing directly over the 
square, Fig. 10. 

Fig. 11. It is here required to mark two unequal 
pitches on two sides of a square bar—then to have a 
piece of plank or board cut so that it shall exactly fit 
both pitches on the bar. The inclination of plank may 
be assumed as A B, and height of both pitches as HA; 
let E F be the seat from which all measurements are 
taken, and transferred to lines that are on the surfaces 
of plank and square with AB; thus giving points to 
direct in drawing line C D; and D E produced. 

Could we apply the bevel J to points C and K, and 
have plumb lines on the edge of plank, then by cutting 
through these and those already marked on the sur¬ 
face, the problem would at once be solved, by making 
both upper and under surface of plank fit the two 
pitches as required. But in practice this would be in¬ 
admissible on account of the great waste of material. 

The proper method is to take any point, E, and cut 
through the plank square with E D, and at C, cut 
through the plank square with CD; here it will be 
noticed that line AD on the surface makes a very 
different pitch to that of A B on the edge of plank, and 
that C D differs from both. 

To understand these points thoroughly is the true 
secret of the nicest element in the joiner’s art—hand 
railing. It being clear that two bevels are required 
one for each pitch—proceed to find them. Make N L 
equal one side of the square. Take N as center, and 
strike an arc touching the line E D; with same radius, 
and L centre, make the intersection in S; join it at L, 


PRACTICAL SOLUTIONS 145 



Fig. 11. 












146 


MODERN CARPENTRY 


which gives bevel R for joint at E. Again take N as 
centre, and strike an arc touching line D C; with same 
radius, and L centre, make the intersection in W; join 
it and L, which gives bevel P for joint at C; the cuts 
being made by these bevels. 


PANELLED CEILINGS IN WOOD AND STUCCO 
BRACKETS, AND SIMILAR WORK. 

It is no part of the duty of this work to show designs 
for wooden ceilings, but in order to illustrate the 
method or methods, of constructing a wooden ceiling 
I deem it proper to show a design in this style the better 
to convey to the student the reason for the various 
steps taken to reach the desired result. 

Fig. 12 shows a section of a roof and the mode of 
constructing a bracketed ceiling under it, but with 
slight modification the same arrangement can be 
adopted for ceilings under floors. The rafters A A are 
18 inches from centre to centre. On these, straps a a, 
3X1^ inches, are nailed at 16 inches apart, and similar 
straps a a, 1*4X1 inch, are nailed to the ceiling joists. 
To the straps are nailed the brackets b b b, shaped to 
the general lines of the intended ceiling, and also 
placed 14 inches apart. The laths are nailed to the 
brackets for the moulded parts, and to straps for the 
flat parts of the ceiling. The brackets and the straps 
for ceiling should not be more than one inch in thick¬ 
ness, for where the brackets and straps occur the 
plaster cannot be pressed between the laths to form a 
key. If the brackets and straps are made thicker, 


PRACTICAL SOLUTIONS 


147 

































































































148 


MODERN CARPENTRY 


parts of the ceiling are apt to be weak and irregular. 
The lathing should be well bonded and have no end 
joints longer than twelve inches on the ceiling, and 
twenty-four inches on the Avails and partitions. No 
joints of laths should be OA^erlapped, as the plaster 
would thereby be made thinner at a part where it 
forms a no key, and Avould thus be liable to crack from 
the vibration of roof, floor or other causes. 

Fig. 13 is a section on a larger scale at right angles 
to that sho\A T n in Fig. 12. Here the ceiling bracket b 
is shown affixed by hooks to the wall, and to the ceiling- 
joists by the strap a a. The ceiling having been plas¬ 
tered, and the mouldings of the cornice run on the 
lathed brackets prepared to receive them, the plaster 
enrichments marked e d e f g are then applied. No. 3 
on Fig. 2 shows a curtain-box in section with its cur¬ 
tain-rod. 

Fig. 14 is a section through a Avindow-head, meeting- 
rails and sill of a window. The safe lintel is placed 
about 10 inches above the daylight of the window, for 
the purpose of allowing Venetian blinds to be drawn up 
clear of the window, and leave the light unobstructed. 
The framing of the window is carried up to the lintel, 
and between it and the upper sash a panel is set in, 
and the blinds hang in front of it. 

Fig. 15 is a cross vertical section, and Fig. 16 a plan, 
looking up, of a skylight on the ridge of a roof, suitable 
for a staircase or corridor. The design can be adopted 
to suit various widths. The skylight is bracketed for 
plaster finishings, in the same manner as the ceiling 
already described. The framing and mouldings at b 
are carried down the side of the light at the same slope 


PRACTICAL SOLUTIONS 



^ N?3. 









































































































150 


MODERN CARPENTRY 


as the sash, till they butt against the sill and bridle 
d and e, forming a triangular panel having for its base 
the cornice c, which is carried round the aperture 
horizontally, and finishing flush with the ceiling, per¬ 
mits the cornice on the corridor to be continued with¬ 



out interruption. Observe in Pig. 16 the lower cornice 
finishes the walls, and the upper mouldings 0 marked 
c in Pig. 15 are carried round the well of the skylight. 

Pig. 17 shows a plan of a panelled ceiling. The" lower 
members or bed mouldings are carried round the walls 
of the room, and tend to build together, and give an 


























PRACTICAL SOLUTIONS 


151 


appearance of support to the several parts of the ceil¬ 
ing. The best way of making panelled ceilings is to 
cover the floor with boarding, and lay down the lines 
of the ceiling on the temporary floor thus formed. 
Then build and lath the ceiling on these lines. When 
it is completed it will be quite firm, and can be cut into 
sections suitable for being lifted up and attached to 
the joisting. The same lines will serve as guides for 
the plasterer setting the moulds for running the cor¬ 
nices and for preparing the circular ornament. 



Fig. 18 shows in plan and section a centre suited for 
this ceiling. By fixing the central portion of it one 
inch from the surface of the finished ceiling and con¬ 
necting it by small plaster blocks placed about one 
inch apart, it forms an excellent ventilator. A zinc 
tube can be led from this centre into a vent, and an 
















































































152 


MODERN CARPENTRY 


air tight valve put on the tube to prevent a down draft 
when the vent is not in use. F is an iron rod fixed to 



Fig. 18. 


a strut or dwang between the joists for the purpose 
of securing a gas pendant or electrolier. 

Let CAB (Fig. 19) be the elevation or the bracket 
of a core, to find the angle-bracket. 

























PRACTICAL SOLUTIONS 153 

First., when it is a mitre-bracket in an interior angle, 
the angle being 45°: divide the curve CB into any 



number of equal parts 12 3 4 5, and draw through the 
divisions the lines Id, 2e, 3f, 4g, 5c, perpendicular to 

























154 


MODERN CARPENTRY 


A B, and cutting it in d e f g c; and produce them to 
meet the line D E, representing the centre of the seat 
of the angle-bracket: and from the points of intersec¬ 
tion h i k 1 c draw lines h 1, i 2, k 3, 14, at right angles 
to D E, and make them equal—h 1 to d 1, i 2 to e 2, 
&c.; and through F1 2 3 4 5 draw the curve of the 
edge on the bracket. The dotted lines on each side 
of D E on the plan show the thickness of the bracket, 
and the dotted lines ur, vswt, show the manner of 
finding the bevel of the face. In the same figure is 
shown the manner of finding the bracket for an obtuse 
exterior angle. Let D I K be the exterior angle: bisect 
it by the line I G, which will represent the seat of the 
centre of the bracket. The lines I H, m 1, n 2, o 3, p 4, 
c 5, are drawn perpendicular to I G, and their lengths 
are found as in the former case. 

To find the angle-bracket of a cornice for interior 
and exterior, otherwise reentrant and salient, angles. 

Let AAA (Fig. 20) be the elevation of the cornice- 
bracket, E B the seat of the mitre-bracket of the in¬ 
terior angle, and H G that of the mitre-bracket of the 
exterior angle. From the points A k a b c d A, or 
wherever a change in the form of the contour of the 
bracket occurs, draw lines perpendicular to Ai or 
D C, cutting A i in e f g h i and cutting the line E B 
in E 1 m n o B. Draw the lines EG, G L B H, and H K, 
representing the plan of the bracketing, and the par¬ 
allel lines from the intersection lmno, as shown 
dotted in the engraving; then make B F and H I per¬ 
pendicular to E B and G H respectively, and each 
equal to i A, o u to h d, n t to g c, m s to f b, 1 r to e a, 
1 p to e k, and join the points so found to give the con- 


PRACTICAL SOLUTIONS 155 

tours of the brackets required. The bevels of the face 
are found as shown by the dotted lines xvyw, &c. 



To find the angle-bracket at the meeting of a concave 
wall with a straight wall. 

Let A D E B (Fig. 21) be the plan of the bracketing 
on the straight wall, and P M G E the plan on the eir- 






















156 


MODERN CARPENTRY 


cular wall, CAB the elevation on the straight wall, 
and GMH on the circular wall. Divide the curves 



C B, G H into the same number of equal parts; through 
the divisions of C B draw the lines C D, 1 d h, 2 e i, &c., 
perpendicular to A B and through those of G H draw 

















PRACTICAL SOLUTIONS 


157 


the parellel lines, part straight and part curved 1 m h, 
2 n i, 3 o k, &c. Then through the intersections hikl 
of the straight end curved lines draw the curve D E, 
which will give the line from which to measure the 
ordinates hi, i 2, k 3, &c. 



Fig. 22 shows the method of finding the angle-bracket 
at the meeting of coves of equal height but unequal 
projection. The height CB is equal to GH, but the 
projection BA is greater than HI. 
























158 


MODERN CARPENTRY 


Fig. 23, Nos. 1, 2, 3, shows the curb and ribs of a 
circular opening (CBA, No. 2), cutting in on a slop¬ 
ing ceiling. No. 1 is a section through che centre B D, 
No. 2 and E F I, No. 3. The height L K is divided into 



Fig. 23. 


equal parts in e, f, g, h, i, and the same heights are 
transferred to the main rib in No. 2 at A, 1, 2, 3, 4, 5, B. 
Through the points A, 1, 2, 3, 4, 5, in No. 2 lines are 
drawn parallel to the axis BEI; and through the 
points e ? f, g, h, i. in No. 1 lines are drawn parallel to 































































PRACTICAL SOLUTIONS 159 

the slope KH. The places of the ribs 1, 2, 3, 4, 5, in 
the latter, and their site on the plan. No. 3, and also 
the curve of the curb, are found by intersecting lines 
m the manner with which the student is already ac¬ 
quainted. 



ON NICHES. 

To describe a spherical niche on a semi-circular plan. 

The construction of this (Fig. 24) is precisely like 
that of a spherical dome. The ribs stand in planes, 
which would pass through the axis if produced. They 
are all of similar curvature. No. 2 shows an elevation 






































160 


MODERN CARPENTRY 


of the niche, and No. 3 the bevelling of the ribs a, b, 
^gainst the front rib at D on the plan; a b is the bevel 
of a, and b c of b. 

Let HBK (Fig. 24) be the plan. It is obvious that 
the ribs mp, nr, will be parts of the quadrant GF 
(No. 3). Transfer the lengths lo, mp, nr, and r s to 
the line G E, as shown at opr s, and raise perpendicu¬ 
lars from these points to the quadrant; G p is the rib 
m p, and o p is the bevel ; G r is the rib nr, and r s is 
the section of the front rib at the crown; the vertical 
projection of the upper arris of this rib will be a semi¬ 
circle with radius ss or s H. 

The niche of which both the plan and elevation are 
segments of a circle. 

No. 25 is the elevation of the niche, being the seg¬ 
ment of a circle whose centre is at E. No. 1, A B C, is 
the plan, which is a segment of a circle whose centre 
is I). Having drawn on the plan as many ribs as are 
required, radiating to centre D, and cutting the plan 
of the front rib in a, be, de; then through the centre 
D draw the line GTI parallel to AC; and from D de¬ 
scribe the curves ml, AG, C H, cutting the line GH; 
and make D F equal to E 0, No. 2. From F as a centre 
describe the curves 1 p 1 and GIH for the depth of 
the ribs; and this is the true curve for all the back 
ribs. 

To find the lengths and bevel of the ribs:—From the 
centre D describe the quadrant and arcs a f, b g, eg, 
d h, &c., and draw f f, g g, h h perpendicular to D H, 
cutting the curve 1 p 1, and the lines of intersection will 
give the lengths and bevels of the several ribs. 


PRACTICAL SOLUTIONS 




Fig. 25. 




























































162 


MODERN CARPENTRY 



Fig. 26. 


























































PRACTICAL SOLUTIONS 163 

Let D in the plan (No. 1, Pig. 26) be the centre of 
the segment. Through D draw E P parallel to A C, 
and continue the curve of the segment to E P. Then 
to find the curve of the back ribs .-—From k 1 m n, any 
points in the curve of the front rib (No. 2) let fall per¬ 
pendicular to the line A B, cutting it in a b c d. Then 
from D as a centre describe the curves a e, b f, c g, e h, 
d h, and from the points where they meet the line E P 
draw the perpendiculars e k, f 1, g m, h n, h o, and set 
up on e k the height ek of the elevation and the cor¬ 
responding heights on their other ordinates, when 
k 1 m n o will be the points through which the curve 
of the radial ribs may be traced. The manner of find¬ 
ing the lengths and bevels of the ribs is shown at 
t u u v v. 

A niche semi-elliptic in plan and elevation. 

Let No. 1, Pig. 27, be the plan, and No. 2 the eleva¬ 
tion of the niche. The ribs in this case radiate from the 
centre D, and with the exception of m g (which will 
be the quadrant of a circle) they are all portions of 
ellipses, and may be drawn by the trammel, as shown 
in No. 4, which gives the true curve of the rib marked 
d in No. 2 and b D in No. 1. The rib c, in the eleva¬ 
tion, is seen at a D in No. 1; the bevel of the end h i is 
seen at A a in No. 3, and that of the end e f at b c. 

To draw the ribs of a regular octagonal niche. 

Fig. 28.—Let No. 1 be the plan, and No. 2 the eleva¬ 
tion of the niche. It is obvious that the curve of the 
centre rib H G will be the same as that of either half 
of the front rib AG, F G. In No. 3, therefore, draw 
A B C D E, the half-plain of the niche, equal to 
A B C H G, No. 1, and make D G E equal to half the 


164 


MODERN CARPENTRY 




1 » -f ** 

Elliptical Niche on Elliptical Plan 
Fig. 27. 




















































PRACTICAL SOLUTIONS 


165 




front rib. Divide D G into any number of parts 12 3 4, 
&e., and through the poihts of division draw lines par¬ 
allel to A G, meeting the seat of the centre of the angle 



















































166 


MODERN CARPENTRY 




Fig. 29. 


























































PRACTICAL SOLUTIONS 


167 


rib C E in iklmno. On these points raise indefinite 
perpendiculars, and set up on them the heights al in 
i 1, b 2 in k 2, and so on. The shaded parts show the 
bevel at the meeting of the ribs at G in No. 1. 

To draw the ribs of an irregular octagonal niche. 

Fig. 29. Let No. 1 be the plan, and No. 2 the eleva¬ 
tion of the niche. Draw the outline of the plan of the 
niche at ABCDEF (No. 3), and draw the centre 
lines of the seats of the ribs B G, HI, &c.; draw also 
GLF equal to the half of the front rib, as given in the 
elevation No. 2, and divide it into any number of parts 
12 3 4. Through the points of division draw dl, c 2, 
b 3, a 4, perpendicular to GF, and produced to the 
seat of the first angle rib G E. Through the points of 
intersection draw lines parallel to the side E D of the 
niche meeting the second angle rib DG; through the 
points of intersection again draw parallels to D C, and 
so on. The curve of the centre rib is found by setting 
up from n o p q G the heights d 1, c 2, &c., on the par¬ 
allel lines which are perpendicular to K G. The curve 
of the rib B G or E G is found by drawing through the 
points of intersection of the parallels perpendiculars 
to the seat of the rib, and setting upon them, at 
h m r I G, the heights dl, c2, &c. No. 4 shows the rib 
C G, and No. 5 the intermediate rib TII. 

DOUBLE CURVATURE WORK. 

To Obtain to Soffit Mould for marking the veneer 
(see Fig. 35), divide the elevation of lower edge of the 
head (Fig. 30) into a number of equal parts, as A, B, 
C, D, E, S, and drop projectors from these points into 


168 


MODERN CARPENTRY 


the plan cutting the chord line A" S" in A", B", C", 
D", E", S". Draw the line s' s', Fig. 35, equal in length 
to the stretch out of the soffit in the elevation (the 
length of a curved line is transferred to a straight 
one by taking a series of small steps around it with 
the compasses, and repeating a like number of the 
straight line), and transfer the points ABC, &c., as 
they occur thereon, repeating them on each side of the 
centre line. Erect perpendiculars at the points, and 
make each of these lines equal in length to its cor- 
despondingly marked line in the plan, as A' a' a' Fig. 
35, equal to A" a a Fig. 32, these letters referring re¬ 
spectively to the chord line and the inside and outside 
edges of the head. Draw the curves through the points 
so found. As will be seen by reference to Fig. 35, the 
mould is wider at the springing than at the crown; 
this is in consequence of the pulley stiles being parallel. 
If they were radial their width would be the same as 
the width of the head at the crown, and the head would 
be parallel; the gradual increase in width from the 
crown to the springing is also apparent in the sash- 
head and the beads, as indicated by the line 0 0, Fig. 
35, which is the inside of the sash-head, and the out¬ 
side of the parting head; this variation in width ren¬ 
ders it impossible to gauge to a thickness from the face 
or the groove in the head from its edges. 

To Form the Head. Having prepared the cylinder 
(Fig. 33) to the correct size, prepare a number of 
staves to the required section, which may be obtained 
by drawing one or two full size on the elevation, as 
shown to the right in Fig. 36. The staves should be 
dry straight-grained yellow deal, free from knots and 


PRACTICAL SOLUTIONS 


169 


FIg * 30 * Fig. 31. 


















































































MODERN CARPENTRY 

O) 

sap, and not be so wide that they 
require hollowing to fit. If the 
veneer is pine, it will probably bend 
dry, but hardwood will require soft¬ 
ening with hot water. One end 




should be fixed as shown in Fig. 34, 
< by screwing down a stave across it. 
Then the other end is bent gently 
over until the crown is reached, 



Fig. 34. 

when another stave is screwed on, 
and the bending continued until 
the veneer is well down all round, 
and a third stave secures it until 
it is thoroughly dry, when the re- 





























PRACTICAL SOLUTIONS 


171 


mainder may be glued on. It is as well to interpose a 
sheet of paper between the cylinder and the veneer, in 
case any glue should run under, which would then ad¬ 
here to the paper instead of the veneer. The head 
should not be worked for at least twelve hours after 
glueing. If a band saw is at hand, the back should be 
roughly cleaned off and the mould bent round it, the 
shape marked, and the edges can then be cut vertically 
with the saw, by sliding it over the cylinder sufficiently 



for the saw to pass. When cut by hand, the mould 
is applied inside and the cut is made square to the face, 
the proper bevel being obtained with the spokeshave, 
and found by standing the head over its plan and 
trying a set square against it. When fitting the head 
to the pulley stiles the correctness of the joints is tested 
by cutting a board to the same sweep as the sill, with 
























172 


MODERN CARPENTRY 


tenons at each end, and inserting it in the mortises for 
the pulleys, as shown at B, Fig. 36. A straight-edge 
applied to this and the sill will at once show if the 
head is in the correct position, and if the edges are 
vertical as they should be. 

To Obtain a Developed Face Mould. Make the line 
i h, Fig. 37, equal to the stretch out of the plan of the 
face of sash-heads, viz., IH, Fig. 32. Transfer the di¬ 
visions as they occur, and erect perpendiculars thereon. 



Fig. 37. 


Make these equal in height to the corresponding ordi¬ 
nates over the springing line in the elevation, and draw 
the curves through the points so found. The groove 
for the parting bead can be marked by running a 
% in. piece around the inside of the sash-head, this 
bead being generally put in parallel. It is sometimes 
omitted altogether, the side beads being carried up 
until they die off on the head. 







PRACTICAL SOLUTIONS 


173 


To Find the Mould for the Cot Bar. Divide its cen¬ 
tre line in the elevation into equal parts, as 0, P, Q, 
R, T. Drop projectors from these into the plan, cut¬ 
ting the chord line 11 in o, p, q, r, t. These lines should 
be on the plan of the top sash, but are produced across 
the lower to avoid confusion with the projectors from 
the other bars. Set out the stretch out of the cot bar 


M R' Q' P o' P' Q' R' T' 

Fig. 38. 

on the line M' O' T', Fig. 38, and erect perpendiculars 
at the points of division, and make them equal in 
length to the correspondingly marked lines in the plan. 
The cot bar is cut out in one piece long enough to form 
the two upright sides as well as the arch. The straight 
parts are worked nearly to the springing, and the bar 
wdiich is got out wider in the centre is then steamed 
and bent around a drum, and afterwards cut to the 
mould (Fig. 38) and then rebated and moulded. The 
arched bar should not be mortised for the radial bars, 
but the latter scribed over it and screwed through from 
inside. 

To Find the Mould for the Radial Bars. Divide the 
centre line of the bar into equal parts, as 1, 2, 3, 4, 5, 
Fig. 30, and project the points into the plan, cutting 
the chord line J J in 1', 2', 3', 4', 5'. Erect perpendicu¬ 
lars upon the centre line from the* points of division, 
and make them equal in length to the distance of the 
corresponding points in the plan from the choFd line, 
and draw the curve through the points so obtained. 











174 


MODERN CARPENTRY 


The other bar is treated in the same manner, the 
projectors being marked with full lines in the plan. 


Fig. 39. 



The soft moulds for the head linings are obtained in 
similar manner to those for the head mould, the width 
being gauged from the head itself 


























































PRACTICAL SOLUTIONS 


175 


A Frame Splayed Lining with Circular Soffit as 

shown in part elevation in Fig. 39, plan Fig. 40, and 
section Fig. 41. The soffit stiles are worked in the solid 



in two pieces joined at the crown and springings. The 
rails are worked with parallel edges, their centre lines 
radiating from the centre of the elevation. Edge 















































1,76 


MODERN CARPENTRY 


moulds only are required for the stiles, and a developed 
mould for the panels. The method of obtaining these 
has been shown on a separate diagram, Fig. 42. 



Method of Obtaining Moulds for Circular Splayed 
Linings. 

Fig. 42. 

To Obtain the Face Moulds for the Soffit Stiles. 
Draw the line EE, Fig. 42, and produce the faces of 













PRACTICAL SOLUTIONS 


177 


the jambs until they meet in point C. From C draw 
the line CD perpendicularly to EE. This line will 
contain the centres of the various moulds, which are lo¬ 
cated by producing the plans of the edges of the stile 
across it e, 1, 2, 3 being the respective centres, and the 
inside and outside faces of the jambs affording the 
necessary radii for describing the arcs A a and B b. 

To Apply the Moulds. Prepare the stuff equal in 
thickness to the distance between the lines e 1 and 2 3, 
Fig. 42. Apply the mould A to the face of the pieces 
intended for the front stile, and cut the ends to the 
mould, and square from the face. Set a bevel as at F, 
Fig. 42, and apply it on the squared ends, working 
from the lines on the face, and apply the mould a at 
the back, keeping its ends coincident with the joints, 
and at the points where the bevel lines intersect the 
face. The piece can then be cut and worked to these 
lines, and the inside edge squared from the face. The 
outside edge is at the correct bevel, and only requires 
squaring slightly on the back to form a seat for the 
grounds. The inside stile is marked and prepared sim¬ 
ilarly. 

The Development of the Conical Surface of the Soffit 

is shown on the right hand of the diagram, Fig. 42, and 
is given to explain the method of obtaining the shape 
of the veneer, but it is not actually required in the 
present construction, as the panel being necessarily 
constructed on a cylinder, its true shape is defined 
thereon, and its size is readily obtained by marking 
direct from the soffit framing when the latter is put 
together. Let the semi-circle E D E, Fig. 42, represent 
a base of the semi-cone, and the triangle E C E its ver- 


178 


MODERN CARPENTRY 


tical section. From the apex C, with the length of 
one of its sides as radius, described the arc E F, which 
make equal in length to the semi-circle E D E by step¬ 
ping lengths as previously described. Join F to C, and 
E C F is the covering of the semi-cone. The shape of 
the frustum, or portion cut off by the section line 
of the linings, is found by projecting the inner edge of 
the lining upon the side C E, and drawing the con¬ 



centric arc I J; then E IJ E represents the covering 
of the frustum. Any portion of this, the panels for 
instance, is found in the same way. To draw the rails, 
divide their centre lines equally on the perimeter, and 
draw lines from the divisions to the centre as HC; 
make the edges parallel with these lines. 

A Window With a Splayed Soffit and Splayed Jambs 
is shown in part elevation, Fig. 43, plan Fig. 44; and 





































PRACTICAL SOLUTIONS 


179 


section Fig. 45. The linings are grooved and tongued 
together, as shown in the enlarged section, Fig. 46. To 
obtain the correct bevel required for the shoulder of 
the jamb and the groove in the soffit, the lining must 



Fig. 44. 


be revolved upon one of its edges until it is parallel 
with the front, when its real shape can be seen. This 
operation is shown in the diagram on the right-hand 



half of the plan and elevation. Draw the line a b, rep¬ 
resenting the face,of the jamb in plan, at the desired 
angle. Project the edges into the elevation, and in¬ 
tersect them by lines c d, projected from the top and 
bottom edges of the soffit in section. This will give the 
projection of the linings in elevation. Then from point 

























180 


MODERN CARPENTRY 


a as centre, and the width of the lining a b as radius, 
describe the arc b b', bringing the edge b into the 
same plane as the edge a. Project point b' into the 
elevation, cutting the top edge of the soffit produced in 
c'. Draw a line from c' to d, and the contained angle 
is the bevel for the top of the jamb. When the soffit 
is splayed at the same angle as the jambs, the same 
bevel will answer for both; but when the angle is 
different, as shown by the dotted lines at e, Fig. 45, 
then the soffit also must be turned into the vertical 
plane, as shown at e' and a line drawn from that point 
to intersect the projection of the front edge of the 
jamb in F; join this point to the intersection of the 
lower edges, and the contained angle is the bevel for 
the grooves in the soffit. (See Fig. 46.) 



The Enlarging and Diminishing of Mouldings. The 

design of a moulding can be readily enlarged to any 
desired dimensions by drawing parallel lines from its 
members, and laying a strip of paper or a straight-edge 
of the required dimensions in an inclined direction be¬ 
tween the boundary lines of the top and bottom edges; 












PRACTICAL SOLUTIONS 


181 


and at the points where the straight-edge crosses the 
various lines, make marks thereon which will be points 
in the new projection, each member being increased 
proportionately to the whole. Projections drawn at 
right angles to the former from the same points will 
give data for increasing the width in like manner. 

To Diminish a Moulding. The method to be ex¬ 
plained, which is equally applicable to enlargement, is 
based upon one of the properties of a triangle, viz., if 
one of the sides of a triangle is divided into any 
number of parts, and lines drawn from the divisions 
to the opposite side of the triangle, any line parallel 
with the divided side will be divided in corresponding 
ratio. See Fig. 47, where ABC is an equilateral tri¬ 
angle, the side AB being divided into six equal parts, 
and lines drawn from these to the apex C. The two 
lines DE and FH, parallel to A B, are divided into 
the same number of parts, and each of these parts bears 
the same ratio to the whole line that the corresponding 
part bears to AB, viz., one-sixth; the application of 
this principle will now be shown. Let it be required to 
reduce the cornice shown in Fig. 48 to a similar one 
of smaller proportions. Draw parallel projectors from 
the various members to the back line A B, and upon 
this line describe an equilateral triangle. Draw lines 
from the points on the base to the apex, then set off 
upon one of the sides of the triangle from C a length 
equal to the desired height of the new cornice as at 
G or H, and from this point draw a line parallel to the 
base line. At the points where this line intersects the 
inclined division lines, dra>v horizontal projectors cor¬ 
responding to the originals. To obtain their length 


182 


MODERN CARPENTRY 


or amount of projection, draw the horizontal line b E, 
Fig. 48, at the level of the lowest member of the cor¬ 
nice, and upon this line drop projectors at right angles 
to it from the various members. Describe the equilat¬ 
eral triangle b i E upon this side, and draw lines from 
the divisions to the apex i. To ascertain the length 



that shall bear the same proportion to b E that the line 
G H bears to A B, place the length of b E on the line 
B A from B to F, and draw a line from F to C: the 
portion of the line G H cut off from J to H is the pro¬ 
portionate length required. Set this length off parallel 
to bE within the triangle, as before described, and 
also draw the horizontal line LH, Fig. 49, making i£ 





















































PRACTICAL SOLUTIONS 


183 


equal in length to a a, Fig. 48. Upon this line set off 
the divisions as they occur on a a, noting that their 
direction is reversed in the two figures. Erect per¬ 
pendiculars from these points to intersect the previ¬ 
ously drawn horizontals, and through the intersections 
trace the new profile. The frieze and architrave are 
reduced in like manner, M B, Fig. 48, representing the 
height of the original architrave, and N TI the reduc¬ 
tion. The cornice can be enlarged similarly by pro¬ 



ducing the inclined sides of the triangle, as shown by 
the dotted lines on Fig. 48, sufficiently to enable the 
required depth to be drawn within it parallel to A B. 
One member has been enlarged to indicate the method, 
which should be clear without further explanation. 

Raking Mouldings. Fig. 50 shows the method of 
finding the true section of an inclined moulding that 
is required to mitre with a similar horizontal moulding 








184 


MODERN CARPENTRY 


at its lower end, as in pediments of doors and windows. 
The horizontal section being the more readily seen, is 
usually decided first. Let the profile in Fig. 50 repre¬ 
sent this. Divide the outline into any number of parts, 
and erect perpendiculars therefrom, to cut a. horizontal 
line drawn from the intersection of the back of the 
moulding with the top edge, as at point 7. With this 
point as a centre and the vertical projectors from 1 to 7 
as radii, describe arcs cutting the top of the inclined 
mould, as shown. From these points draw perpendicu¬ 



lars to the rake, to meet lines parallel to the edges of 
the inclined moulding drawn from the corresponding 
points of division in the profile, and their intersections 
will give points in the curve through which to draw 
the section of the raking mould. When the pediment 
is broken and a level moulding returned at the top, its 
section is found in a similar manner, as will be clear 
by inspection of the drawing. If the section of the 
raking moulding is given, that of the horizontal mould 
can be found by reversing the process described above. 
Fig. 51 shows the application of the method in finding 
the section of a return bead when one side is level and 









PRACTICAL SOLUTIONS 


185 


the other inclined, as on the edge of the curb of a sky¬ 
light with vertical ends. 

Sprung Mouldings. Mouldings curved in either ele¬ 
vation or plan are called “ sprung/’ and when these 
are used in a pediment, require the section to be deter¬ 
mined as in a raking moulding. The operation is sim¬ 
ilar to that described above up to the point where the 
back of the section is drawn perpendicular to the in¬ 
clination, but in the present case this line E x, Pig. 52, 



is drawn radiating from the centre of the curve, and 
the projectors are drawn parallel to this line. The 
parallel projectors, 1, 2, 3, 4, 5, are also described from 
the centre until they reach the line E x, when per¬ 
pendiculars to this line are raised from the points of 
intersection to meet the perpendicular projectors. 

Mitreing Straight and Curved Mouldings Together. 
If a straight mitre is required, draw the plan of the 
mouldings, as in Fig. 53, and the section of the straight 
mould at right angles to its plan as at A. Divide its 




186 


MODERN CARPENTRY 


profile into any number of parts, and from them draw 
parallels to the edges intersecting the mitre line. From 
these intersections describe arcs concentric with the 
plan of the curved mould, and at any convenient point 
thereon draw a line radial from the centre. Erect 
perpendiculars on this line from the points where the 
arcs intersect it, and make them equal in height to the 
corresponding lines on the section of the straight 
moulding A, and these will be points in the profile of 



the curved moulding B. "When it is required that the 
section of both mouldings shall be alike, a circular 
mitre is necessary, and its true shape is obtained as 
shown in Fig. 53. Draw the plan and divide the profile 
of the straight moulding as before, drawing parallels 
to the edge towards the seat of the mitre. Upon a line 
drawn through the centre of the curved moulding set 
off divisions equal and similar to those on the straight 
part, as 1 to 8 in the drawing. From the centre of the 




















PRACTICAL SOLUTIONS 


187 


curve describe arcs passing through these points, and 
through the points of intersection of these arcs with 
the parallel projectors, draw a curve which will be the 
true shape of the mitre. Cut a saddle templet to this 
shape, and use it to mark the mouldings and guide the 
chisel in cutting. 



To Obtain the Section of a Sash Bar Raking in Plan. 

Fig. 54 represents the plan of a shop front sash with 
bars in the angles. On the left hand is shown the sec- 

























188 


MODERN CARPENTRY 


tion of the stile or rail into which the bars have to 
mitre. Divide the profile of moulding into a number 
of parts, as shown from 1 to 6, and from these points 
draw parallels to the sides of the rails intersecting the 
centre line of the bar. Also draw perpendiculars from 
the same ponts to any line at right angles to them, as 
at A. Draw a line at right angles to the centre line of 
the bar, and on it set off the divisions from 1 to 6 as 
at A. Draw projectors from these points parallel to 
the centre line of the bar, and where they intersect 
the correspondingly numbered lines drawn parallel 
with the sides of the sash will be points in the curve of 
the section of the bar. It will be noticed that there is 
no fillet or square shown on the bar, and that in trans¬ 
ferring the points from the line A they must be re¬ 
versed on each side of the centre. Should a fillet be 
required on the bar, additional thickness must be given 
for the purpose. Three methods of forming the rebates 
in the bar are shown, the screwed saddle beads being 
the best for securing the glass. 

Interior Shutters Include Folding, Sliding, Balanced, 
and Rolling. Exterior Consist of Hanging, Lifting, 
Spring, and Venetian Shutters. 

FOLDING, or as they are frequently called, BOX¬ 
ING SHUTTERS, because they fold into a boxing or 
recess formed between the window frames and the 
walls, are composed of a number of narrow leaves, 
framed or plain, as their size may determine, rebated 
and hinged to each other and to the window frames. 
They should be of such size that when opened out they 
will coyer the entire light space of the sash frame and 
a margin of a % in. in addition. Care must be taken 


PRACTICAL SOLUTIONS 


189 


to make them parallel, or they will not swing clear 
at the ends; and as a further precaution, they should 
not be carried right from soffit to window board, but 
have clearance pieces interposed at their ends about 
% in. thick, dhe outer leaf, which is always framed, 
is termed a shutterj the others are termed flaps. It is 
not advisable to make the shutters less than 1*4 in. 
thick, and flaps over 8 in. wide should be framed; those 
less than 8 in. may be solid, but should be mitre 
clamped to prevent warping. In a superior class of 
work the boxings are provided with cover flaps which 
conceal the shutters when folded, and fill the void when 
they are opened out. The sizes and arrangements of 
the framing are determined by the general finishings 
of the apartment, but it is usual to make the stiles of 
the front shutters range with those of the soffit and 
elbow linings. When Venetian or other blinds are used 
inside, provision is made for them by constructing a 
block frame from 2 */ 2 to 3 in. thick inside the window 
frame and the shutters are hung to this. 

The leaves are hung to each other with wide hinges 
called back flaps, that screw on the face of the leaves, 
there not being sufficient surface on the edges for butt 
hinges. In setting out the depth of boxings, at least 
V 8 in. should be allowed between each shutter to pro¬ 
vide room for the fittings; the shutters are fastened 
by a flat iron bar hung on a pivot plate fixed on the 
inner left-hand leaf, and having a projecting stud at 
its other end which fits into a slotted plate, and it is 
kept in this position by a cam or button. Long shutters 
are made in two lengths, the joint coming opposite the 
meeting rails of the sashes; -these are sometimes re¬ 
bated together at the ends. 


T 


190 


MODERN CARPENTRY 



Fig. 55. 











































































































PRACTICAL SOLUTIONS 


191 


Pig*. 55 is a sectional elevation of a Window Fitted 
with Boxing Shutters having a cover flap and spaces 
for a blind and a curtain. One-half of the elevation 
shows the shutters opened out and the front of the 
finishings removed, showing the construction of the 
boxings, &c. The plan, Fig. 56, is divided similarly, 
one 'half showing the shutters folded back, with por^ 



tions of soffit cornice, &c.; the other half gives the plan 
and sections of the lower parts, the dotted lines show¬ 
ing the window back. 

Fig. 57 is a vertical section and Fig. 58 an enlarged 
section through the boxings. The framed pilaster cov¬ 
ering the boxing is cut at the level of the window 
board, and hung to the stud A', this being necessary 
for the cover to clear the shutters when open (see 
dotted lines on opposite half). The cover flap closes 
into rebates at the top and bottom, as shown in sec- 

















































192 


MODERN CARPENTRY 



tion in Fig. 55. The window 
back is carried behind the 
elbows, and grooved to receive 
the latter. The rails of the 
soffit must be wide enough to 
cover the boxings, and should 
have the boxing back tongued 



into it, as shown in the section, 
Fig. 55. When the linings of 
an opening run from soffit to 
the floor uninterruptedly, they 
are called jamb linings; but 
when they commence, as in 
the present instance, under the 


























































































































PRACTICAL SOLUTIONS 


193 


window board, they are termed elbow linings, the cor¬ 
responding framing under the window being the win¬ 
dow back. 

Sliding Shutters are used instead of folding shutters 
in thin walls, and consist of thin panelled frames run¬ 
ning between guides or rails fixed on the soffit and 
window board, which are made wider than usual for 



Fig. 59. 


Fig. 60. 


that purpose. When open, they lie upon the face of 
the wall adjacent to the opening. They are so seldom 
used that they do not call for illustration. 

Balanced or Lifting Shutters are shown in section in 
Fig. 61, and plan in Fig. 62\ They consist of thin pan¬ 
elled frames, the full width of and each half the height 









































M 


MODERN CARPENTRY 



of the window, hung with weights in a cased frame in 
t* similar manner to a pair of sashes. The frame ex¬ 
tends the whole height of the window, and is carried 
down behind the floor joists to the set off. The win- 






































































PRACTICAL SOLUTIONS 


195 



Fig. 62. 







































































































































196 


MODERN CARPENTRY 


dow board is hinged to the front panelling or shutter 
back, so that when the former is lifted, the shutters 
can pass down behind the back out of sight. Cover 
flaps hung to the outer linings on each side close over 
the face of the pulley stiles and hide the cords. When 
it is desired to close the shutters, the cover flaps are 
opened out flat, as shown to the left of the plan. The 



Fig. 63. 


window board can then be lifted, and the shutters run 
up, a pair of flush rings being inserted in the top edge 
of each for that purpose. The window board is next 
shut down, the inside shutter brought down upon it, 
the outside one pushed tight up to the head, and the 
meeting rails, which overlap an inch, fastened with a 
thumb-screw. The bottom rail of the upper shutter 
is made an inch wider than the other, for the pur¬ 
pose of showing an equal margin when overlapped. 
Square lead weights have generally to be used for 


















PRACTICAL SOLUTIONS 197 

these shutters in consequence of their comparative 
heaviness. 

Rolling’ or Spring 1 Shutters are made in iron, steel 
and painted or polished woods, and though somewhat 
monotonous in appearance, are, in consequence of their 
convenience in opening and closing, and in the case 
of the metal kinds, the additional security against fire 
and burglary, fast superseding all other kinds, both 
for internal and external use, especially in shops and 
public buildings. They consist, in the case of the 
wood varieties, of a series of laths of plano-convex, 
double convex, oval, or ogee section, fastened together 
by thin steel or copper bands passing through mortises 
in the centre of the thickness, as shown in Fig. 63, 
and secured at their upper ends to the spring con¬ 
tainer. These metal bands are supplemented by sev¬ 
eral waterproof bands of flax webbing glued to the 
back sides of the laths. The upper edge of each lath 
is rounded, and fits into a corresponding hollow in 
the lower edge of the one above it; this peculiar over¬ 
lapping joint, whilst preventing the passage of light, 
etc., between the laths, readily yields when the shutter 
is coiled around the barrel. The spring barrel is usually 
made of tinned iron plate, and encases a stout spiral 
spring wound around an iron mandrel with squared 
ends to which a key is fitted for winding the spring 
up. The ends of the mandrel project from the case 
and are fixed in bracket plates at each end of the 
shutter—in the case of shop fronts, in a box or recess 
behind the fascia; one end of the spring is secured to 
the barrel, the other end to the mandrel; the shutter is 
secured to the barrel by the metal bands mentioned 


198 


MODERN CARPENTRY 



Fig. 84. 










































PRACTICAL SOLUTIONS 


199 


above, and the normal condition of things is that when 
the shutter is coiled up the spring is unwound. The 
pulling down of the shutter winds up the spring, and 
the tension is so arranged that it does not quite over¬ 
come the weight of the shutter and the friction when 
the shutter is down, so that the latter must be assisted 
up with a long arm. Great care should be taken in 
fixing these shutters to arrange the barrel perfectly 
level and parallel with the front, and to securely fix the 
brackets. These may be bolted to the girder or breast- 
summer, or screwed to the fixings plugged in the wall, 
as shown by dotted lines in Fig. 64. Where possible a 
wood groove should be formed on each side of the open¬ 
ings for the shutters to work in, but iron 
channels (Fig. 65) are frequently used. 

These are cemented into a chase in the 
wall or pilaster. Fig. 64 is a section 
through a shop fascia, showing shutter 
and blind barrels fixed to the face of the Fi s- 65 - 
wall, where no provision has been made beneath the 
girder. Fig. 66 shows the adaption of one of these 
shutters to the inside of a window; the barrel is fixed 
in a seat formed beneath the window, the top of which 
is hinged, to gain access to the coil. The minimum 
space required for a coil, for a shutter about 6 ft. high 
is 101/2 in. A friction roller F should be fixed 
close to the back rail to prevent chafing as the coil un¬ 
winds; the shutter is lifted by means of a flush ring 
in the L iron bar at the top, and this ring engages with 
a tilting hook in the soffit to keep the shutter up. The 
metal varieties of these shutters are wound up by the 
aid of balance weights or bevel wheel gearing. 









200 


MODERN CARPENTRY 



Fig. 66. 























PRACTICAL SOLUTIONS 


201 


WINDOWS GENERALLY. 

Size and Position. The size and position of win¬ 
dow openings are influenced by the size of the rooms, 
and the purposes for which the building is used. For 
the sake of ventilation, and also to secure good light¬ 
ing, the windows should be placed at as great a height 
as the construction of the room will allow. In dwell¬ 
ing-houses the height of the sill is usually about 2 ' 6 " 
above the inside floor level. 

Construction. The framework holding the glass of 
the window may be fixed or movable. It must be so 
prepared that the glass can be replaced easily when 
necessary. In warehouses, workshops and similar 
buildings, the frames holding the glass are often fixed 
as Fast Sheets (Fig. 66 ^ 2 )- As, however, this arrange¬ 
ment affords no means of ventilation, it is more usual to 
have the glass fixed in lighter frames called Sashes. If 
the sashes are hung to solid rebated frames, and open 
as doors do, the windows are called Casement Sashes. 
If they slide vertically and are balanced by weights or 
by each other, the window is a Sash and Frame Win¬ 
dow. Other methods of arranging sashes, either 
hinged, pivoted or made to slide past each other, are 
described in detail later. 

Sashes. The terms used for the various parts of 
sashes and fast sheets are somewhat similar to those 
employed in describing doors. Thus, the Styles are the 
outer uprights, and the Rails are the main horizontal 
cross-pieces: top rails, meeting rails, and bottom rails 
being distinguished. Any intermediate members, 
whether vertical or horizontal, are named Bars. 


202 


MODERN CARPENTRY 



Fig. 66K. 

Sashes are from l 1 /^ to 3 inches thick. The inner edge 
of the outer face is Rebated to receive the glass. The 
inner face is left either square, chamfered, or moulded; 



Fig * 67 - Fi &- 68. Fig. 69. Fig. 70. 

two common forms of moulding are lamb Vtongue (Fig 
68) and ovolo (Fig. 69). The size of the rebate is indi- 





























PRACTICAL SOLUTIONS 


203 


cated in Fig. 70; it varies with the thickness of the sash, 
its depth being always a little more than one-third this 
thickness. The width of the rebate varies from a quar¬ 
ter of an inch to half an inch, and the mould is usually 
sunk the same depth as the rebate. This last fact is of 
some importance, as it affects the shoulder lines; and 
with hand work it influences the amount of labor in 
the making of the sashes. 

As little material as possible is used in the sashes, in 
order that the light shall not be interfered with. In gen¬ 
eral, the styles and top rail are square in section before 
being rebated and moulded. In casement sashes, how¬ 
ever, it is often advisable to have the outer styles a 
little wider than the thickness, especially when they are 
tongued into the frame. The width of the bottom rail 
is from one and a half to twice the thickness of the 
sash. Sash bars which require rebating and moulding 
on both sides, should be as narrow as possible, in order 
not to interrupt the light. They are usually from 
five-eighths of an inch to one and a quarter inches wide. 

Joints of Sashes. The sashes are framed together by 
means of the Mortise and Tenon Joint (Fig. 71). The 
proportions of the thickness and width of tenons, 
haunched tenons, &c., are to a large extent appli¬ 
cable here. Hardwood cross-tongues are sometimes in¬ 
serted to strengthen the joints while thick sashes 
should have Double Tenons. An alternative to halving 
in sash bars is to arrange that the bar which is sub¬ 
jected to the greater stress—as for example, the ver¬ 
tical bars in sliding sashes, and the horizontal bars in 
hinged casement sashes—shall be continuous; this con¬ 
tinuous bar is mortised to receive the other, which is 


204 


MODERN CARPENTRY 


scribed i. e. cut to fit the first, and on which the short 
tenons are left. This method is called Franking the 
Sash Bars, and is illustrated in Fig. 72. 




Fig. 72. 


Casement Windows. Casement windows may be 
hinged in such a manner that they open either inwards 
or outwards. They may consist either of one sash, or of 
folding sashes, and are hung with butt hinges to solid 
rebated frames. These Frames consist of jambs, head 





























PRACTICAL SOLUTIONS 


205 


and sill. The head and sill “run through M and are 
mortised near the ends to receive tenons formed on the 
ends of the jambs. The upper surface of the sill is 
weathered to throw off rain water. Casement windows 
which reach to the floor are usually called French Case¬ 
ments. - Their sashes require an extra depth of bottom 
rail. 

Casement Sashes Opening Inwards. Figs. 73, 74 show 
the elevation and vertical and horizontal sections, of 
a window opening in a 14" brick wall fitted with a 
casement window having folding sashes to open in¬ 
wards. In this class of window the frame is rebated for 
the sashes on the inner side. Each sash has, on the 
outer edge of the outer style, a semi-circular tongue, 
which fits into a corresponding groove in the jamb of 
the frame. This tongue renders the vertical joint be¬ 
tween the sash and frame more likely to be weather 
proof; it is to provide for the tongue that the extra 
width of style already referred to is necessary. The 
tongue, however, is often omitted, as in Fig. 77. It 
will be seen readily that, if the sash were in one width, 
it would be impossible to have a tongue on more than 
one edge of it. With casement sashes opening inwards 
the greatest difficulty is found, however, in making a 
water-tight joint between the bottom rail of the sash 
and the sill of the frame. Figs. 77 and 78 show two 
methods by which this may be accomplished. An es¬ 
sential feature of all these sashes is a small groove or 
Throating on the under edge of the bottom rail; this 
prevents the water from getting through. The groove 
in the rebate of the sill (Pig. 78) is provided to collect 
any water that may drive through the joint. This 


206 


MODERN CARPENTRY 




Fig. 76. 




































































































PRACTICAL SOLUTIONS 207 

water escapes through the hole bored in the center of 
the sill. 



When casement sashes are hung after the manner of 
folding doors, the vertical joint between the meeting 
styles is rebated. Alternative methods of rebating 




are shown in Pig. 79 and '80. Fig. 80 is known as a 
Hook Joint and is the better one. 
































208 


MODERN CARPENTRY 


Casement Sashes Opening Outwards. These are more 
easily made weatherproof than inward-opening sashes. 
The chief objections to their adoption are that they are 
not easily accessible for cleaning the outside, especial¬ 
ly in upper rooms and that they are also liable, when 
left open, to be damaged by high winds and to let in 
the rain during a storm. Fig. 81 is a sketch of one 
corner of such a window. It wdll be noticed that these 



frames, like door frames, have the exposes arrises 
moulded in various ways, and that the sashes may 
either be hung flush with one face of the frame (as in 
Figs. 76 and 77), or fit in the thickness of the frame 
(Figs. 78 and 81). The sill shows to be Double Sunk, 
i. e., to have the upper surface—upon which the bot¬ 
tom rail of the sash fits—rebated with two slopes 
(weatherings). 





















PRACTICAL SOLUTIONS 


209 


Sash and Frame Window. In this class of window, 
which is by far the most common, because it is easily 
made weatherproof, there are Two Sashes, which slide 
past each other in vertical grooves, and are usually 
balanced by iron or leaden weights. As will be seen 
from Fig. 82 the frames form cases or boxes in which 
the weights are suspended. They are hence called 



Cased Frames. Pulley Styles (Fig. 86) take the place 
of the solid rebated jambs of casement windows. The 
pulley styles, Outside and Inside Linings, and Back 
Lining (Fig. 82) together form a box which is sub¬ 
divided by a vertical Parting Slip suspended as shown 
in Fig. 82. In superior window frames of this kind, the 
pulley styles and linings are tongued and grooved to¬ 
gether as shown in Fig. 83. In commoner work the 
tongues and grooves are often omitted. The frame 








































210 


MODERN CARPENTRY 


must be so constructed that the sashes can be removed 
easily for the purpose of replacing broken sash-lines. 
To enable this to be done, the edge of the inside lining 
is either made flush with the face of the pulley style 
(Fig. 83), or it is rebated slightly as shown in Fig. 83. 
The edge of the outside lining projects for a distance 
of about three-quarters of an inch beyond the face of 
the pulley style, to form a rebate against which the 
outer (upper) sash slides. The outer sash is kept in 



position by the Parting Lath (Fig. 82) which fits into 
a groove in the pulley style. The groove for the inner 
(lower) sash is formed by the parting lath and a 
Staff Bead or Stop Bead which is secured by screws. 
The staff bead on the sill is often made from two to 
three inches deep to allow the lower sash to be raised 
sufficiently for ventilation at the meeting rails without 
causing a draught at the bottom (Fig. 83). 

. A vertical section through the head of the frame is 
similar to a horizontal section across the pulley style. 







































PRACTICAL SOLUTIONS 


211 


except that the back lining and parting slip are of 
course absent. 

The sill of the frame is solid and weathered, and 
should always be of hardwood, preferably oak or teak. 
The sill has a w T idth equal to the full thickness of the 
frame. When the weathering has two steppings, it is 
known as a Double Sunk Sill. An alternative to the 
plan of having the width of the sill of the full thickness 
of the frame, is to arrange At so that the outside edge 
is flush with the outside face of the bottom sash, as 
shown in Fig. 89. With a sill arranged in this manner, 
and double sunk, there is less danger of water driving 



Fig. 84. 


through the joint between the sash and the sill than 
with a sill the full thickness of the frame. In order 
to render watertight the joint between the wooden 
and stone sills of window frames, a metal tongue is 
often fixed into corresponding grooves cut into the un¬ 
der side of the wooden sill and the upper surface of 
the stone sill. A rebated joint between the two sills 
serves the same purpose as the metal tongue. 

Fig. 84 shows the methods of fixing the pulley style 
into the head and sill respectively, when the width of 
the sill is equal to the full thickness of the frame. The 
Pulleys on which the sash', lines run—sash or axle pul¬ 
leys are fixed in mortises near the upper ends of the 






212 


MODERN CARPENTRY 


pulley styles. It is also necessary to have a removable 
piece in the lower part of each pulley style, to allow 
of access to the weights. This piece is named the 
Pocket Piece. It may be cut as shown in Fig. 85; its 
position is then behind the lower sash, and it is hidden 



from view when the window is closed. Or, the pocket 
piece may be in the middle of the pulley style as shown 
in Fig. 84; the vertical joints between the pocket piece 
and the pulley style are then V shaped to prevent dam¬ 
age to the paint in case of removal. 



<S Removable- pocket piece, 


Pulley Style\ 



Fig. 85. 


Sashes. The only difference between the joints of 
sliding sashes and those of the casement sashes already 
described is in the construction of the meeting rails. 
Each of the meeting rails is made thicker than the sash 
to the extent of the thickness of the parting lath; other¬ 
wise there would be a space between them equal to the 



































PRACTICAL SOLUTIONS 


213 


thickness of the parting lath. The joint between them 
may be rebated or splayed. The angle joints between 
the ends of the sash styles and the meeting rails are 




often dovetailed as shown in Fig. 86. They are, how¬ 
ever, stronger if the styles are made a little longer, 
the projecting part being moulded, and mortise and 




















214 


MODERN CARPENTRY 


tenon joints used as shown in Figs. 86 and 87. The 
projecting ends of the styles are called Joggles; they 
assist in enabling the sashes, especially in wide win¬ 
dows, to slide more freely. When as is usually the case, 
both sashes slide and are balanced by weights, the 



Fig. 87. 

window is known as a Double-Hung sash and frame 
window. If one sash only slides, and the other is fixed 
m the frame, the window is Single-Hung. Figs. 78 to 
80 show the details of a sash and frame window fixed 
in a one-and-a-half-brick-thick wall and having a stone 
head and sill. 




































































PRACTICAL SOLUTIONS 


215 


For the sake of appearance, or when it is required to 
have wider windows than can be arranged with one 
pair of sashes, two or three pairs of sashes are often 
constructed side by side in the same frame. When 
three pairs of sashes are used, it is usual to have the 
middle pair wider than the others; such a combination 
(Fig. 73) is named a Venetian Window. The vertical 
divisions between adjacent pairs of sashes are called 
Mullions. These mullions may be constructed in sev¬ 
eral different ways. If the middle pair of sashes only is 
required to slide, the mullions may be solid from l 1 /^" 
to 2" thick, and the sash-cord conducted by means of 
additional pulleys to the boxes, which are at the outer 
edges of the frame. Figs. 84 and 86 show this arrange¬ 
ment. If it is desirable to have all the sashes to slide, 
the mullions must be hollow to provide room for the 
weights. Figs. 85 and 87 show details of a mullion with 
provision made for one weight to balance the two 
sashes adjacent to it. With this arrangement the sash- 
cord passes round a pulley fixed into the upper end of 
the weight. If stone mullions are used in the window 
opening, separate boxings may be made so that each 
pair of sashes is hung independently as shown in Fig. 
83, and the window becomes, as it were, two or three— 
as the case may be—separate window frames, with the 
sill and head each in one length for the sake of 
strength. 

The Hanging of Vertical Sliding Sashes. As shown 
in illustrations already given, the sashes of sash and 
frame windows are balanced by cast-iron or leaden 
weights. The best cord is' employed for hanging sashes 
of ordinary size, while for very heavy sashes the sash 


216 


MODERN CARPENTRY 


lines are often of steel or copper. The staff bead and 
parting bead having been removed, the cords are passed 
over the axle pulleys (which are best of brass to pre¬ 
vent corrosion) and are tied to the upper ends of the 
weights. The weights are passed through the pocket 
holes and suspended in the boxes. The pocket pieces 
having been replaced, the upper sash, which slides in 
the outer groove, is hung first, the free ends of the 
cords being either nailed ,into grooves in the outer 
edges of the sash or secured by knotting the ends after 
passing them through holes bored into the styles of the 
sash. The upper' sash having been hung, the parting 
laths are fixed into the grooves in the pulley styles, and 
the lower (inner) sash is hung in a similar manner, 
after which the staff beads are screwed in position. 
Care should be taken to have the cords of the right 
lengths; if the cords for the upper sash are too long 
the weights will touch the bottom of the frame, and 
cease to balance the weight of the sash before the lat¬ 
ter is closed. If the cords for the lower sash are too 
short, the weights will come in contact with the axle 
pulleys, and thus prevent it from closing. Several dif¬ 
ferent devices for hanging sashes—the objects of 
which are either to render unnecessary the use of 
weights or to facilitate the cleaning of the outside of 
the window—have been patented, and are in more or 
less general use. A detailed description of these is, 
however, beyond the scope of this book. 

Bay Windows. A bay window is one that projects 
beyond the face of the wall. The side lights may be 
either splayed or at right angles to the front. The 
window openings may be formed by having stone or 


PRACTICAL SOLUTIONS 


217 


brick mullions or piers at the angles, against which 
the window frames are fixed, or the wooden frame¬ 
work of the window may be complete in itself. When 
the latter is the case, it is usual to have stone or brick 



work to the sill level as shown in Fig. 88. Bay win¬ 
dows naturally lend themselves to decorative trep f - 
ment. With the addition of masonry or brickwork the\ 
often assume a massive and bold appearance. When 































































































218 


MODERN CARPENTRY 


usually by a wooden cornice, and the wooden roof is 
covered with lead, slates or tiles. The window frames 
may be arranged as fixed lights, sash and frame, or 



casements. The most usual arrangement is to have 
the lower lights fixed, and the upper ones as sashes 
hinged to open for ventilating purposes. Figs. 88 to 90 











































































































































PRACTICAL SOLUTIONS 


219 


show the details of a bay window with splayed side 
lights, the upper side lights being hinged on the tran¬ 
som to open inwards. 

Windows With Curved Heads. When a window 
opening is surmounted by an arch, the top of the win¬ 
dow frame requires to be off the same curvature as the 
under side (soffit) of the arch. In the case of fixed 



sashes, or of solid frames with casement sashes, the 
head of the frame is “cut out of the solid.” A head 
which, owing to the size of the curve, cannot easily be 
obtained in one piece, is built up of segments, the 



joints being radial to the curve, and secured by hard¬ 
wood keys. As an alternative method, the head may 
be built up of two thicknesses—with overlapping 
joints—and secured together by screws. 

































220 


MODERN CARPENTRY 


A sash and frame window in such an opening may 
have only the outside lining cut to the curve of the 
arch, the inner side of the frame being left square. The 
upper sash will then require a top rail with a straight 
upper edge and a curved lower edge, as shown in Fig. 
91. 



When the head of the frame has to be curved, it 
may 

(1) be built up of two thicknesses with overlapping 
joints, and secured by screws; it may 

(2) be formed of three thicknesses of thin material, 
bent upon a block of the correct radius, and well glued 
and screwed together; or 

(3) the head may be of the same thickness as the 
pulley styles, with trenches cut out of the back (up¬ 
per) side, leaving only a veneer on the face-side under 
the trenches. Wooden keys are glued and driven into 
the trenches after the head has been bent upon a block 
to the required shape. 

A strip of stout canvas glued over the upper side 
will strengthen the whole materially. The outside and 















PRACTICAL SOLUTIONS 


221 


inside linings are in such a case cut to the required 
curvature, and when nailed in position hold the head 
in shape. The end joints of the linings may have hard¬ 
wood cross-tongues. 

Shop Windows. The main object in view in the con¬ 
struction of shop windows is to admit the maximum of 
light, and to give opportunity for an effective display 
of the goods. The glass is in large sheets, and there¬ 
fore is especially thick to secure the necessary strength. 
Shop windows are usually arranged as fast sheets, with 
provision for ventilation at the top. The glass is held 
in position by wooden fillets, and is fixed from the in¬ 
ner side. The chief constructional variations are found 
in the pilasters, cornice, provision for sign-board, sun- 
blind, and the arrangement of the side windows. Figs. 
92 to 95 show the details of a typical example. 

The Fixing of Window Frames. Window frames 
may be built into the wall—which has usually a re¬ 
cessed opening to receive them—as the brickwork pro¬ 
ceeds, or they may be fixed later. In the former case, 
the ends of the sill and head project and form Horns, 
which are built into the brickwork and help to secure 
the frame. Wooden bricks or slips may also be built 
into the wall, the frames being nailed to them. 

In the latter case, the frames are secured by wooden 
Wedges, which are driven tightly between the frame 
and the wall. These wedges should be inserted only at 
the ends of the head and sill and directly above the 
jambs; otherwise the frame might be so strained as to 
interfere with the sliding of the sashes. Window 
frames as well as* door frames should be bedded 
against a layer of hair-mortar placed in the recess. 


222 


MODERN CARPENTRY 



Fig. 94. 







































































































PRACTICAL SOLUTIONS 



223 


Linings. When window frames are not of sufficient 
thickness to come flush with the inner face of the wall, 
the plaster may be returned round the brickwork and 

















































224 


MODERN CARPENTRY 


finished against the frame, or a narrow fillet of wood 
may be scribed to the wall and nailed to the frame. In 
dwelling-houses, however, the more usual way is to 
fix linings similar to those used for outer door frames. 
The width of the linings depends upon the thickness 
of the wall; they should project beyond the inner face 
of the wall for a distance equal to the thickness of the 



plaster, and are usually splayed so that they will not 
interfere with the admission of light. The inside of 
window and door openings usually are finished similar¬ 
ly; thus, the Architrave, or Band Moulding, which is 
secured to the edge of the linings and to rough wooden 
Grounds is fixed along the sides and top in both cases. 


























PRACTICAL SOLUTIONS 225 

The bottom of the window opening is finished with a 
Window Board which is tongued into the sill of the 
frame. The board is about 1^4 inches thick, and is 
made wide enough to project beyond the surface of the 
plaster for a distance of about 18 inches. The project¬ 
ing edge is nosed (rounded) or moulded. It is longer 
than the opening, to allow the lower ends of the archi¬ 
trave to rest upon it. 

When the walls are thick, the linings are often 
framed and panelled. Such linings may terminate on 
a window board at the sill level, or the inner side of 
the wall may be recessed below the sill level and the 
linings carried to the floor as shown in Fig. 96. 

Window Shutters. Although not used to the same 
extent as formerly, wooden window shutters are fitted 
occasionally to close up the window opening. Window 
shutters, which are arranged generally on the inner 
side of the window, may be hinged as box shutters, or 
may be vertically sliding shutters. 

Box Shutters consist of a number of leaves or nar¬ 
row frames which are rebated and hinged together, an 
equal number being on each side of the window open¬ 
ing, the outer ones on each side being hung to the win¬ 
dow frame. When closed they together fill the width 
of the window-space, and when open they fold behind 
each other so that the front one forms the jamb lining 
of the window frame. If the walls are thick, the shut¬ 
ters can be arranged to fold in the thickness of the 
wall; if the wall is a thin one it is necessary to con¬ 
struct projecting boxes into which the shutters fold. 
The nature of the framing of the shutters depends upon 
the surrounding work; it is usual to have the outer sur- 


226 


MODERN CARPENTRY 


face framed and moulded, and the inside finished bead- 
flush. The arrangement of box shutters requires that 
the shutters on the same side shall vary in width so 
that they will fold into the boxes on each side of the 
window, the outermost shutter (which is the widest) 
then acting as the window lining. Fig. 96 shows a hori¬ 
zontal section through one side of a window, showing 
hinged shutters folding so that a splayed lining is ob¬ 
tained. Fig. 97 shows hinged shutters consisting of 
one narrow and one wide shutter on each side of the 



Fig. 97. 


opening. This arrangement is suitable for a thin wall, 
where it is undesirable to have boxes for the shutters 
projecting beyond the face of the wall. For hang¬ 
ing window shutters it is usual to use back-flap hinges; 
the joint at the corner of the shutters in Fig. 97 is 
named a rule joint. 

Sliding Shutters, working in vertical grooves and 
balanced by weights, are sometimes used. They re¬ 
quire that the wall under the window sill shall be re¬ 
cessed; the floor also often needs trimming to allow 



























PRACTICAL SOLUTIONS 


227 


space for them to slide 
sufficiently low. To 
hide the grooves in 
which the shutters 
slide, thin vertical flaps 
are hung to the window 
frame, and the window 
board is also hinged at 
the front edge to allow 
the shutters to slide be¬ 



low the sill. Figs. 98 and 
99 are sections of ver¬ 
tical sliding shutters. 

Hinged Skylights. 

Skylights which are 
hinged to open are fit¬ 
ted upon the upper 
edge of a Curb or 
frame fixed in the plane 
of the roof, the com- 










































































































































228 


MODERN CARPENTRY 


mon rafters being “trimmed” to the required size to 
receive the curb. The curb is made from material l 1 /^ 
to 2 inches thick, and of width such that its upper edge 
stands from 4 to 6 inches above the plane of the roof. 
The Angle Joints of the curb may be dovetailed or 
tongued and nailed. The Sash Frame rests upon the 
upper edge' of the curb; it is from 2 to 2% inches 
thick, and consists of stiles and top rail of the same 
thickness, and a bottom rail which is thinner than the 
stiles by the depth of the rebate. Bars are inserted in 
the direction of the slope of the roof, and the butt 
hinges used for hanging the sash are invariably fixed as 
the under side of the top rail. 

The joints between the curb and the roofing slates or 
tiles are made weatherproof with sheet lead. . At the 
upper end—the back of the curb—a small Lead Gutter 
is formed, with the lead going underneath the slates 
and overlapping the upper edge of the curb. The sides 
of the curb may be flashed with soakers—short lengths 
of sheet lead which are worked in. between the slates— 
or the joint may be made with one strip of lead form¬ 
ing a small gutter down the side of the curb. In either 
case the lead overlaps the upper edge of the curb. At 
the lower end of the curb, the lead overlaps the slates. 
To prevent water from rising between the glass and 
the upper side of the bottom rail, sinkings are cut into 
the rail. (See Fig. 100). 

Dormer Windows. Instead of having the light in or 
parallel to the plane of the roof, it affords a more ar¬ 
tistic treatment of the roof, and often gives a better re¬ 
sult in lighting, if the window is fixed vertically. The 
general arrangement of the framing, as well as of the 


PRACTICAL SOLUTIONS 


229 


sashes, depends upon the kind of roof, the width of the 
window required, and the general style of architecture 
of the building. 

The construction of a dormer window necessitates 
trimming of the rafters, and the arrangement of pro¬ 
jecting framework, the front of which consists of Cor¬ 



ner Posts and Crossrails—rebated to receive hinged 
sashes—which are connected to the main roof by other 
crossrails and by Braces. This framework is surmount¬ 
ed by a Roof which may be either ridged, or curved 
outline, or flat. By arranging a ridged roof to over- 








230 


MODERN CARPENTRY 


hang, and adding suitable Barge Boards and Finial 
(Fig. 101) a dormer window may be made to improve 
the general appearance of the roof of a building. The 
sides of the dormer may be either boarded and covered 
with the same kind of material as the roof, or they may 
be framed for sidelights. 


cfi 



As dormer windows are generally in exposed posi¬ 
tions, and the sashes are arranged as casements to open, 
their efficiency depends largely upon the perfection of 
the joints between the sashes and the frame. It ought 
to be mentioned however, that with sashes hung fold¬ 
ing, semicircular tongues on their hanging stiles are 
by far the best. Figs. 101 and 102 give the details of 




































PRACTICAL SOLUTIONS 


231 


a dormer window, with sidelights, fixed in a roof of 
ordinary pitch. The sashes, which are hnng folding, 
open inwards. The roof may be boarded and covered 
with lead, or it may be covered with slates or tiles. 
The joints between the roofing slates of the main roof, 
and the roof and sides of the dormer, are made weath¬ 
erproof with sheet lead flashings. Figs. 103 and 104 
show a dormer window fixed in a Mansard roof; in this 
example there are no side lights. 



Fig. 103. 


Fig. 104. 


Large Skylights and Lantern Lights. For lighting 
the well of a large staircase, or a room which, for some 
reason, cannot be lighted with side windows, specially 
large skylights are often necessary. These are of a 
more elaborate construction than the skylights already 
described; they vary considerably in size, shape and 
design; the plan may be rectangular, polygonal, cir- 




















232 


MODERN CARPENTRY 


cular, or elliptical, and the outline may be pyramidal, 
conical, or spherical. The framework may be of either 
wood or iron. To support such a skylight, a strong 
wooden curb is framed into the roof, and projects from 
6 to 9 inches above the roof surface. The joints be¬ 
tween the curb and the roof are made watertight with 
sheet lead. The framework of the skylight may con¬ 
sist of rebated quartering; with separate lights which 
fit into the rebates of the framing; or the sashes them- 



Fig. 105. 


Fig. 106. 


selves may be constructed with strong angle stiles, 
which are mitred together, and provided with either a 
hardwood tongue inserted in the joint, or with a 
wooden roll on top to keep out the water. 

With skylights of this description, channels for con¬ 
densed water should always be provided. These are 
placed at the upper inner edge of the curb, the re¬ 
mainder of the inside face of the curb being covered 
by either panelled framing or match boarding. 






























PRACTICAL SOLUTIONS 


233 


Figs. ^.05 and 106 give details of a skylight having 
the form of a square pyramid. In this example the 
four triangular lights are mitred at the angles, and 
have wooden rolls over the joints. Figs. 107 and 108 
show elevation and part plan of a skylight with a 
curved roof surface. 

A Lantern Light differs from the skylights just de¬ 
scribed in having, in addition, vertical Sidelights. The 
sidelights consist of sashes, which, by being hinged or 




Fig. 108. 


pivoted, are often available for ventilation. As they 
are in exposed positions, the greatest care is required 
in order to obtain watertight joints. When the side¬ 
lights are hinged on the bottom rail, they open in¬ 
wards ; when on the top rail they open outwards. When 
they are hung on pivots, the pivots are fixed slightly 
above the middle of the sash. Figs. 109 and 112 show 
details of a rectangular opening surmounted by a lan¬ 
tern light which is hipped at both ends, and has side¬ 
lights arranged to open inwards. 
















234 


MODERN CARPENTRY 


The construction of skylights and lantern lights af¬ 
fords good examples of the application of geometry 
to practical work as described in previous pages. When 



Section through C.D. 

Fig. 109. 


the roof-lights are pyramidal as shown in Figs. 106 and 
110, and a separate frame is constructed as shown in 



Fig. 110. 


Fig. 110, the methods of obtaining the lengths and 
bevels of the hip rafters are similar to those described 
for getting hip rafters. When the roof-lights mitre 






































































PRACTICAL SOLUTIONS 


235 


against one another, the sizes of the lights and the 
bevels of the angle stiles which mitre together are ob¬ 
tained as shown at X in Fig. 112. With lights of 



curved outline, the shapes of the hip rafters or angle- 
stiles, as well as the developed surfaces, are obtained 
as explained before. 

Lay-Lights. At the ceiling level of roof-lights used 
for staircase wells, or in similar positions, it is often 



Fig. 112. 



Fig. 113. 


Fig. 114. 



considered advisable, for the sake of appearance, to 
have a horizontal second light called a lay-light. This 
consists of a sash—or if the space is large, a number of 
sashes—fixed into frames in the ceiling. The chief 



















































236 


MODERN CARPENTRY 


feature of lay-lights is in the attempt at decoration by 
arranging the bars in some ornamental design (Figs. 
113 and 114). The lay-lights are often glazed with 
ornamental glass, which, although it improves the ap¬ 
pearance, diminishes the amount of light transmitted. 

Greenhouses and Conservatories. In this type of 
building which is largely constructed of wood and 
glass, the framework is usually of moulded and re¬ 
bated quartering, with side sashes fixed in the rebates. 
As in the case of skylights, the roof-lights, which in 
this case reach from the ridge to the eaves, have no 
crossbars, since these would impede the flow of water 
running down the slope of the roof. Care should be 
taken to have the bars strong enough to carry the glass 
without sagging; and it is well to remember that when 
a roof is of flat pitch a heavy snowstorm will throw a 
large additional weight upon it, while with a steep 
roof the wind has much power. The distance apart of 
the bars which carry the glass ranges from 12 to 18 
inches, and the lengths of the sheets of glass should be 
as great as possible, so as to diminish the number of 
cross-joints, since these allow of accumulations of dirt 
which cannot be removed easily. These roof-lights are 
constructed in exactly the same manner as skylights; 
they are, however, often much larger, and require to be 
thicker, unless purlins are placed to support them. 
When, as is often the case, part of the roof-light is 
made to open, this part—often a narrow strip at the 
highest part of the roof (Fig. 115)—is made as a sep¬ 
arate light, which overlaps the upper edge of the fixed 
lower light. Additional ventilation is secured by ar¬ 
ranging the side sashes to open. 


PRACTICAL SOLUTIONS 


237 


The above description is intended merely to outline 
the broad principles of the construction of conserva¬ 
tories, but it should be remembered that the details, 
while conforming to casement and roof-light construc¬ 
tion generally, lend themselves to considerable varia¬ 
tion in design and arrangement. 



Fig. 115. 


A Bay Window with solid frame and casement lights 
is shown in Fig. 116 to 118. Two methods of fitting 
such a window, with folding shutters, are given in 
this plan. In the half plan at C, the shutters fold 
into a boxing projecting into the room, and at D they 
fold back upon the face 6f the wall, which is splayed 
to receive them. The sills of the frame are mitred at 



















































238 


MODERN CARPENTRY 


the angles, the joint cross tongued and fixed with a 
handrail bolt, which should be painted with red lead 
before insertion. The joints in the head are halved 
together, the mullions stub tenoned and fixed with 



coach screws. The jambs are tenoned and wedged into 
the head and sill. The transom tenoned into the jambs 
and mullions, and secured with bolts. The mullions 
may be worked in one piece as shown at D, or built 
up as at C, and tongued and screwed together. 






















































































































































PRACTICAL SOLUTIONS 


239 



Flan, shewing' aJUernatiy-e- treatment ofjiidshings 















































































































































240 


MODERN CARPENTRY 


g 

p 

1 J 

(!!■ 1 



.i| 


ij 



oi 



— 


1 if r~ 

IH 


l! 

i 1 i :- -J. 


11 M n 

J 1 




■ 

XT 1 ll’l 


Jnsidc eUvrtioio of half pjan C QutsHe ^Miowrfoir pian ~Q ' >. 


Fig. 119. 


A Cased Frame Bay Window is shown in the half 
plan, half inside elevation, and central vertical section, 
Nos. 1, 2 and 3, Fig. 119. This window is composed of 


































































































































PRACTICAL SOLUTIONS 


241 


three ordinary sash frames, the sills connected at the 
angles either by halving and screwing or by mitreing 
them and fastening the joint with a hand rail bolt. The 
heads are tied together with a short piece of 1 in. stuff 
screwed across the top of the joints, and the joints in 
the linings are covered by the mullions of the blind 
frame B. The latter, made 2 in. wide, forms an en¬ 
closure for Venetian blinds. Boxings are formed in 
the elbows between the sash frames and the interior 
face of the wall, the front of the opening being fin¬ 
ished off with a moulded ground and architrave. These 
form receptacles for the folding shutters, which are 
curved in plan, and when opened out convert the octag¬ 
onal bay into a segmental niche. The window back 
and the seat beneath are also curved to parallel sweeps. 
The window board also follows the sweep, and is re¬ 
bated to receive the shutters, a shaped bead being 
fixed on the soffit to form a stop at the top. Nos. 4, 5 
and 6 on the same plate illustrate another method of 
finishing a bay window. In this case the frame is solid, 
and is fitted with outward opening casement lights. A 
blind frame is provided, and the shutters fold on to the 
face of the jamb and wall, the outer edges passing be¬ 
hind the rebated edges of the architrave; the lat¬ 
ter is continued down to the floor, and elbow linings to 
correspond with the shutters are fitted beneath the 
window board. These are fitted to the window back 
in the manner indicated by the dotted lines in the 
plan No. 4. The section No. 6 shows the treatment of 
the roof of the bay, which is segmental in section and 
covered with shaped pan-tiles. The ribs, which are 
elliptic at the hips, are notched into a wall plate rest- 


242 


MODERN CARPENTRY 



Fig. 120. 


Fig. 121. 

































































































































PRACTICAL SOLUTIONS 243 

ing on the stone cornice, and are nailed at the top into 
a shaped rib fixed on the face of the wall; the ribs are 
covered with weather boarding, which affords a good 
fixing for the tiles. The wall is carried by a breast- 
summer formed of two 12 in. by 6 in. balks bolted to¬ 
gether, with spacing fillets between, and the soffit is 
carried by three brackets fixed to the breastsummer 
and the head of the window frame. 

A Pivoted Light in a solid frame is shown in eleva¬ 
tion in Fig. 122 and section Fig. 123. These are used 
chiefly in warehouses, lanterns, and other inaccessible 
positions, the lights being opened and closed either by 
cords and pulleys or by metal gearing. For small 
lights the frames are usually made out of 4% in. stuff 
by 2 or 3 in. wide. Small lights are pivoted horizontal¬ 
ly, large ones vertically. The pivot should be fixed to 
the frame, not the sash, and from y 2 in. to 1 y 2 in. above 
the centre, according to the weight of the bottom rail. 
The lower part of the sash should exceed in weight the 
upper part, just sufficient to keep it closed; its action 
may be easily demonstrated by inserting two bradawls 
in the stiles, and balancing them on the fingers. The 
sash is inserted and removed from the frame either by 
means of plough grooves in the edges of the stiles, as 
shown by the dotted lines in Fig. 122, or by cutting a 
notch through the face of the stile for the passage of 
the pin, which is concealed when in use by the guard 
beads. This latter is the better method, as it does not 
reduce the strength of the sash, as does the former, by 
cutting away the wedging. The stop beads at the 
sides are cut in two, one payt being fixed to the frame, 
the other to the sash. Their joints can be at any angle 


244 


MODERN CARPENTRY 




' « 

\ X • 



<u 

£ 


-C 

to 

r) 

CO 


b g 

CO 

c 


a 

rH 

X 

bO 

-*—» 

. o 

•M 

> 


s 


cJ 




O 


c 


o 


*-* 


u 


ID 


CO 




G 


ci 


C 

M 

_o 

IN 

tH 

cj 


> 

bo 

<u 




W 




























































PRACTICAL SOLUTIONS 


245 


greater than that made by a line tangent to the sweep 
at the point of intersection a Fig. 124, hut for the pur¬ 
pose of using the Mitre Block, they are generally made 
at an angle of 45 deg. A curved joint has no advan¬ 
tage over a straight one, except in being more expen¬ 
sive. 

To Hang the Sash. Insert the pivots in the frame 
quite level, but do not screw them. Then with the try 
square resting on the top of the pins, square lines 
across the jambs. Then remove the pivots and insert 
the sash, which should be fitted rather tightly at first, 
and square the lines on to the sash. Return these on 
the edges, and keep the edge of the hole in the socket 
plate to the line, and the plate itself in the middle of 
the thickness. After the socket is sunk in, and the 
notches cut, test the sash and correct the joints, which 
should be a bare % in. clear all round. 

To Find the Position to Cut the Beads. After fit¬ 
ting them round, remove them and open the sash to the 
desired angle, which should be less than a right angle, 
so that the water may be thrown off. Lay the beads 
upon the sash upside down for convenience of marking, 
and draw a line along their edge upon the jambs at the 
point where the line meets the faces of the frame; 
square over lines as at a a Fig. 123; the position is 
shown in Fig. 122, the outer dotted lines indicating 
the beads. Next replace the beads and transfer the 
marks to them, cutting them off the mitre block (re¬ 
member that the mark is the longest point of the mitre). 
The upper portions outside and the lower inside are 
fixed to the frame, and these are shaded in the draw¬ 
ings. The remainder of the beads are fixed to the sash. 


246 


MODERN CARPENTRY 


The above describes the method when the sash is 
grooved. Where the beads are slotted, a variation 
must be made with reference to the top cut (see Fig. 



124). In this case the sash must be drawn out and 
rested upon the pin, then the bead laid on it and 
marked as before, the intersection a giving the mitre 
point. 

































PRACTICAL SOLUTIONS 


247 


A Bull’s-Eye Frame with a pivoted sash is shown in 
Figs. 125 and 126 and enlarged detail of the joint. This 
frame is built up in two thicknesses, glued and screwed 
together, each ring being in three pieces breaking joint. 
The beads may be steamed and bent round, or worked 
on the edge of a board that has been cut to the sweep, 



Fig. 126. 


and cut off in two lengths. The sash is made in three 
pieces, with butt radial joints bolted together. To en¬ 
able the sash to open, a plane surface must be provided 
at the centre, equal to the thickness of the sash and 
beads, as shown by the dotted lines. Having fitted the 
sash in, and the beads around each side, brad them tern- 




















248 


MODERN CARPENTRY 


porarily to the sash, lay a straight-edge across it paral¬ 
lel with the centre, and square up with the set square 
a line at each side equal in length to the thickness, then 
cut the pieces so marked off with a fine saw, both beads 
and sash, and glue them to the centre of the frame, and 
fix the pivots to these frames and proceed as in a square 
frame. 



Fig. 127. Fig. 128. Fig. 129. 


Laying Out a Circular Louvre: Suppose the frame 
of the louvre to be formed of four pieces, as shown at 
Pig. 127. These sections may be formed of one thick 
piece of plank, or may be built up of several thick¬ 
nesses. If of one thickness, the joints may be held to¬ 
gether with handrail screws, or dowelled and keyed. If 
of several thicknesses then the joints can be broken or 
overlapped, and the pieces either screwed or nailed 
together. To set out the louvre boards, make a sketch 
of the whole thing, full size, as shown at Pig. 127, then 
set up the section or side as shown at Fig. 128, then 
project from the quick of the bead, then draw a line 
C D, Pig. 129, parallel to the inclination of bevel of 

























PRACTICAL SOLUTIONS 


249 


the louvre boards. This will give the major axis of an 
ellipse. Bisect this line and draw E, F, G, at right 
angles to C, D, making E, F, and F G each, equal to 
half the radius from the centre of the frame to the 
quick of the bead. The ellipse C, E, D, G, may be 
struck by any of the methods shown. As, of course, all 
the louvre boards are at one angle, each forms a por¬ 
tion of the same ellipse. From the centre of the front 
edge of each louvre board, project across to the major 
axis as shown. Now from any of the louvre boards set 
off, at right angles, lines from the upper and lower 
surfaces as shown at H, Fig. 128, then the distance K, 
is the amount of projection of the lower surface in 
front of the upper. Taking half the distance of K, 
measure it off on each side of the centres 1, 2, 3, 4, 5, 6. 
Fig. 129, then through the points last obtained draw 
lines parallel with the minor axis, the upper line rep¬ 
resenting the top front board or arris of the boards, and 
the dotted line the bottom arris. Now measure the dis¬ 
tances K, from C to L, and from D to M, and construct 
the ellipse for the underside of the boards. The fol¬ 
lowing will be found a simple method for making out 
each louvre board: Cut a thin mould equal to one- 
quarter of the ellipse, the edges of the louvre boards 
having been planed to proper bevel, as shown at N, Fig. 
128, square the centre line across the piece that has 
to form the louvre board and lay it on the setting out at 
Fig. 129, the centre line of course corresponding with 
C D. The quarter mould which was prepared can now 
be laid on the louvre board, with its curve standing 
directly over the development, Fig. 129, the face side 
of the material being of course the side marked out. 


250 


MODERN CARPENTRY 


The curves for the underside may be marked off on the 
arrises direct from the development, and the mould 
then applied to the other side, taking care to adjust it 
in the right position and to the marks made. The ends 
should be sawed and turned to the lines. The next 
proceeding will be to set out the frame for the grooves; 
these are represented in the conventional sketches, 4, 5 
and 6. It will be noticed from Fig. 132 that the bottom 
louvre board is not grooved in all round; this is a bet- 



Fig- 130. Fig. 131. F i g . 132 . 


ter method than bringing it out to the front and thus 
destroying a part of the margin bead of the frame. The 
cutting of the grooves and the fitting in of the louvre 
boards requires careful working in order to get good 
joints. It must be clearly understood that Fig. 129 is 
not a full development of all the boards edge to edge, 
as that could not be represented in this space, but 
enough is shown to give a clear idea of how the’lines 
are obtained. The full breadth of each is represented 
by the dimension lines 0, P, R, S, T, U, V, and from 
this all the other is easy. 









PRACTICAL SOLUTIONS 


251 


The Construction of Doors. Doors are named in ac¬ 
cordance with their modes of construction, position, 
style or the general arrangement of their parts, and 
also the method in which they are hung, as Battened, 



Battened Framed and Braced, Panelled or Framed, 
Entrance, Vestibule, Screen, Sash, Diminished Stile, 
Double Margin, Gothic, Dwarf, Folding, Swing, Jib, 
Warehouse Hung, &c. The essentials in the design and 
construction of doors are,'for the first, that they shall 
have a due proportion to the building or place they 

























































252 


MODERN CARPENTRY 


have to occupy and be suitably ornamented; in the sec¬ 
ond, that their surfaces shall remain true and their 
parts be so arranged and connected that their shape 
will be unalterable by the strains of usage and the ef¬ 
fects of weather. The various examples illustrated will 
indicate the points to be considered in designing doors 
for sundry situations, and the methods of construction 
herein described will supply the necessary information 
to meet the constructive requirements. 

Battened and Battened Framed and Braced Doors 
are shown in Figs. 133 and 134. These doors are suit¬ 
able for positions where one or both sides are exposed 
to the weather. Little or no attempt is made to orna¬ 
ment them—economy of cost, strength and utility be¬ 
ing the chief requirements of this class of door, which 
are fitted to coach-houses, W. C.’s and outhouses gen¬ 
erally. 

The plain Battened Door (Fig. 133) is composed of 
battens A, from % to 1*4 in. thick, ploughed and 
tongued in the joints with straight tongues which 
should be painted before insertion, nailed to three 
ledges, B from 1 in. to l 1 /^ in. thick, usually with 
wrought nails long enough to come through and be 
clinched on the back side. The ends of the ledges are 
better fixed with screws, and their top edges as well as 
those of the braces C should be bevelled to throw off 
the water, as shown in the detail, Fig. 135. The lower 
edges may be throated or bevelled under, as shown. 
The braces should be placed so that their lower ends 
are at the hanging side, for if in the opposite direc¬ 
tion, they will be useless to prevent the door racking. 
Their ends should be notched into the ledges about 1 in. 


PRACTICAL SOLUTIONS 


253 


deep and iy 2 in. from the ends, with the abutment 
square to the pitch of the brace. Narrow doors are 
sometimes made without braces, but they seldom keep 
“square.” These doors are hung with wrought-iron 
strap hinges called cross garnets, which should be fixed 
on or opposite the battens, whether placed on the face 
or back of the door. 



VerticalSection ' 
Fig. 136. 



Fig. 137. 


The Framed Battened and Braced Door (Figs. 136 
to 137) differs from the former in that the battens and 
ledges are enclosed on three sides by a frame of a 
thickness equal to the combined thickness of the bat¬ 
tens and ledges, so that it is flush on each side with 




























254 


MODERN CARPENTRY 


them. The boards are tongued into the frame at the 
top and sides, and the ledges are framed into the stiles 
with barefaced tenons. The braces should not be taken 
into the angle formed by the stile and rail, but be kept 
back from the shoulder about 1 in., as shown. If the 
brace is placed in the corner, the strain thrown on it 
has a tendency to force off the shoulder, unless the door 
is very narrow, when the brace will be nearly upright. 
These doors, as in fact all framed work exposed to 
damp, should be put together with a quick drying paint 
instead of glue in the joints, because ordinary glue has 
such an affinity for water that it will soften in damp 
situations releasing its hold, and also be the means of 
setting up dry-rot in the timber. The battens in these 
doors should be made 1/16 in. slack for each foot of 
width to allow for subsequent expansion, or otherwise 
the shoulder will be forced off. The framework of 
these doors is first made and wedges up, then the bat¬ 
tens folded in and driven up into the top rail and 
nailed to the ledges, after which the braces are cut 
tightly in and nailed to the battens in turn, and the 
whole cleaned off together. In large gates of this de¬ 
scription it is usual to stub tenon the braces into the 
rails, in which case they must be inserted first and 
wedges up with the framing. 

Framed or Panelled Doors are of several kinds, dis¬ 
tinguished by the number or treatment of their panels, 
or by the arrangement of the mouldings, as follows: 

Two to Twelve Panel Doors. 

Square and Sunk. When a thin panel is used with¬ 
out mouldings, as shown at A in the elevation diagram 
of a Framed or four-panel door (Fig. 138). 


PRACTICAL SOLUTIONS 


255 



Moulded and Square. When one side of the panel is 
moulded and the other plain, as at B. 

Bead Flush. When one side of the panel is flush, or 
nearly so, with the frame,' and with a bead worked 
round the edges to break the joints, as at C. 









































































25fi 


MODERN CARPENTRY 


Bead Butt. When the bead is worked only on the 
two sides of the panel, as at D. 

Raised Panel. When the centre part of the panel is 
thicker than the margin. There are four varieties of 
raised panels: 


Fig. 140. 


1. The Chamfered. In this the panel is chamfered 
down equally all round, from the centre to the edge 
when square, or from a central ridge if rectangular, as 
shown. 



Plan and Sections of a Chamfered 
Raised Panel. 


Fig. 142. Fig. 141. 


2. Raised and Flat or Raised and Fielded. When a 

chamfer is worked all round the edge, leaving a flat in 
the centre, as at A, Fig. 142. 

3. Raised, Sunk and Fielded (as at B, Fig. 142). 
When the chamfer starts from a marginal sinking be¬ 
low the face. 















PRACTICAL SOLUTIONS 


257 


4. Raised, Sunk and Moulded (as at C, Fig. 143). 
When the edge of the sinking is moulded. 

Stop Chamfered. When the edges of the framing are 
chamfered and stopped near the shoulders. 

Bolection Moulded. When the panel moulding stands 
above, and is rebated over the edges of the framing, as 
shown at Fig. 142. 



Fig. 143. 


Fig. 143 14 . 


Double Bolection Moulded. When the moulding on 
each side of the door is made in one solid piece, grooved 
to receive the panel, and is itself grooved and tongued 
into the framing. This variety is shown at Fig. 142. 

Constructive Memoranda; The outside vertical mem¬ 
bers of doors (in common with all framed work) are 


































258 


MODERN CARPENTRY 


called stiles. The one the hinges are fixed to is called 
the hanging stile, the one containing the lock the strik¬ 
ing stile. In a pair of doors the two coming together 
are called the meeting stiles. The inside vertical mem¬ 
bers are the mountings, or more commonly muntings. 
The horizontal members are rails, respectively, top, 
frieze, middle, or lock, and bottom. The panels are 




Fig. 145. 


named similarly. When the grain of these run hori¬ 
zontally, they are said to be “laying” panels; when 
vertically, “ upright. ” Doors are called Solid-Moulded 
when the moulding is wrought or “stuck” in the sub¬ 
stance of the framing itself, as is shown at Fig. 145; 
and Planted when the moulding is worked sep¬ 
arately and bradded around the frame, as shown at 
a, Fig. 143, and D, Fig. 143^. These are also called 
sunk mouldings. 
















































PRACTICAL SOLUTIONS 259 

Bead Flush Panels are commonly made as shown at 
a, Fig. 146, but such panels will, unless made of thor¬ 
oughly seasoned stuff, inevitably split when drying. 
The correct way to obtain the effect of bead flush pan¬ 
elling is to work the beads upon the edges of the 
framing, as shown in Figs. 147 and 148. 



Bead Butt Panels are better kept about 1/32 in. be¬ 
low the framing, as a truly flush surface is difficult to 
prepare through the yielding of the panel, and when 
produced, will seldom last, as the shrinkage of the 
panel and frame is unequal. Planted-in “sunk” mould- 














































260 


MODERN CARPENTRY 


ings should be fixed to the framing, not to the panels, 
as shown in Fig. 143, D; for if fixed to the panels, when 
the latter shrinks the moulding will be drawn away 
from the frame, leaving an unsightly gap. The back 
edge of the moulding should be bevelled under as 
shown, so that when bradded in, the front edge will 
keep close down to the panel. As it is not permissible 



Fig. 147. 



Fig. 148. 


to brad polished mouldings, except in the case of in¬ 
ferior work, these are usually glued to the frame, and 
their back edges should of course be square. The panel 
should be polished before the moulding is planted in, 
so that in case of shrinkage a white margin will not 
be shown. When, however, the moulding is wide and 
thin, it is unavoidable that it be fixed to the panel to 
keep its front edge down, and to overcome the diffi- 



















































PRACTICAL SOLUTIONS 261 

culty of shrinkage. The stiles in vertical and the rails 
in laying panels are prepared with a second shallow 
groove as shown, and the moulding, made with an extra 
v ide quirk, is cut in and pressed back into the groove. 
It is glued to the panel, and shrinks with it without 
showing a gap. The cross pieces across the ends of the 
panel have their quirks shot off to the proper width, 
and are sprung in after the side pieces are in place. 
These are do welled but are not glued, or when there is 
a moulding at the back, the face side is slot screwed. 

Bolection Mouldings are intended to be fixed to the 
panels, which is rendered necessary by their great 
width and thickness, and they are rebated over the 
edge of the framing to prevent the interstices produced 
by the shrinkage showing. From % to 3/16 in. is suf¬ 
ficient for this purpose, and more should not be given 
or the edge will be liable to curl off. To prevent the 
panel being split by the fastenings when it shrinks, the 
moulding is fixed by slot screws, as shown in Fig. 143, 
the slots being cut across the grain. See the back 
view, Fig. 144. The moulding should also be screwed 
together at the mitres, and the latter may be grooved 
and tongued with advantage, and dropped into the 
panel as a complete frame, where it is fixed as de¬ 
scribed, the heads of the screws being covered by the 
interior moulding, which, if a sunk one, is glued to the 
frame, and if a bolection, dowelled to the panel, or, as is 
sometimes done in inferior work, bradded, and the 
holes filled up with colored stopping. It is not good 
construction to glue the moulding to the panels in any 
case, as the alteration of size in the latter, due to the 
state of the atmosphere, is very liable to cause them to 


262 


MODERN CARPENTRY 


split, if fixed immovably, or when swelling, to disar¬ 
range the mitres. The bradding in of mouldings is less 
likely to do this if the brads are not placed too thickly, 
as they yield slightly to the pressure. Framing is 
usually grooved % in. deep for the panels, and the lat¬ 
ter given y 8 in. play sideways, but fitting close length¬ 
ways of the grain. This is sufficient for panels up to 
2 ft. wide. Over that width the grooves should be % 
in. deep, and the panel enter % in. at each side. Ordi¬ 
nary dry stuff will eventually shrink about % in. to 
the foot, and will swell equally if exposed to damp. 
When wide panels are used they will warp less if glued 
up in several pieces, as the pull of the fibres is lessened 
by the cutting, and the effect of the warping is dimin¬ 
ished in the same ratio as their width. Much can be 
done to ensure the permanent flatness of panels by pay¬ 
ing attention to the way the boards have been cut from 
the tree. The direction of the annual rings on the end 
will indicate this, and the various pieces should have 
their similar sides placed together. What is meant by 
this will be rendered plain by an examination of Fig. 
149. When a panel is glued up with the hollow or heart 
sides of the rings all on one face as at A, and the board 
warps, it will case in one continuous curve, as shown in 
the unshaded diagram, whilst if glued up with the 
heart sides reversed alternately, as shown at B, it will 
assume the serpentine shape shown in the unshaded dia¬ 
gram. Boards cut radically or with the annual rings 
perpendicular to the surfaces, as at C, will swell less 
than the others, and will not warp perceptibly. 

Proportions. The size of doors depends so much upon 
the scale and design of the buildings they occupy, that 


PRACTICAL SOLUTIONS 


263 


no definite data can be given, within reasonable lim¬ 
its, for important doors; but it may be pointed out that 
very large doors not only tend to dwarf a building or 
a room, but they also take up a great deal of space in 
opening, and the difficulty of preserving their accurate 
fitting increases in direct ratio with the size. The 
following may be taken as an indication of the more 
usual dimensions given to ordinary good class dwelling 
house doors: Entrance Doors, from 7 ft. to 8 ft. 6 in. 



illustrating the Effect 
of Position on the Parts 
of a Panel. 

Fig. 149. 

high by from 3 ft. to 4 ft. 6 in. wide, 2 to 2% in. thick. 
Reception Rooms, 7 ft. by 3 ft. 3 in. by 2 in. Bed- 
Rooms, 6 ft. 8 in. by 2 ft. 8 in. by 1 y 2 in. Details of in¬ 
terior doors; stiles and top rails, in common work, out 
of 4!/2 in., muntings and frieze rails 4 in., middle and 
bottom rails 9 in. Superior doors vary much, but gen- 

















264 


MODERN CARPENTRY 


erally stiles and rails are somewhat wider than the 
above, muntings and frieze rails narrower. Height of 
lock rail usually 2 ft. 8 in. to its centre. This is a con¬ 
venient height for the handle, which is generally placed 
in the middle of depth of rail. When an entrance door 
is approached from a step the middle rail is kept about 
6 in. lower, to bring the height of the handle con¬ 
venient. 

Common Doors, both internal and external, are made 
of “yellow pine” or Georgia pine throughout. A bet¬ 
ter class of interior doors have yellow pine frames and 
white pine panels. The latter wood should not be 
used for external work, as it is far too soft and will 
not stand wet. Superior Internal Doors are made 
throughout of Honduras mahogany, black walnut and 
oak; also of pine and bay wood, veneered with Spanish 
mahogany. External Doors of oak, teak, walnut and 
pitch pine. 

In constructing doors of any of the above mentioned 
figure woods, great attention must be paid to the ar¬ 
rangement of the members, so as to balance the figure, 
and this may also well be studied in the conversion of 
the plank. For instance, two stiles, each having pro¬ 
nounced figure at one end, and the other end plain, 
should have the figured ends placed at the bottom. This 
gives the effect of solidity, whilst the reverse would 
make the door look top heavy. Similarly the upper 
rails should be plain, the lower figured. It must be 
understood the above only applies when the wood is a 
mixed lot. When the wood is handsomely figured 
throughout, the point of most importance is the effect 
of its position upon the figure, and this is so great that 


PRACTICAL SOLUTIONS 


265 


in some of the light-colored woods, wainscot and bay- 
wood for instance, a piece that in one position will ap¬ 
pear richly figured will in another show quite plain 
and dull. The best way to judge the effect is to prop 
the pieces up in the approximate positions they will oc¬ 
cupy when finished, - facing a top light. Then when 
standing a few feet off, the play of light on the fibres 
will be observed. Deep-colored woods, such as teak 
and Spanish mahogany, may have their figure brought 
out by slightly oiling them, which will facilitate their 
arrangement. Panels also require balancing, the more 



Fig. 150. 


heavily marked ones being placed below plainer ones, 
and symmetrically arranged, either in pairs, or figured 
in the centre and plain outside. In all cases where the 
figure is coarse, taking a truncated elliptic shape, the 
base or wider part should be kept downwards. The 
panel at B is upside down from an artistic point of 
view. This arrangement is known in the workshop as 
“placing the butts down,” although as a matter of 
fact the width of the figure is not due to its being 
towards the butt end of the tree, but merely to the ac¬ 
cidental position the surface of the board occupies with 
relation to the annual rings, which are more or less 




















266 


MODERN CARPENTRY 



Pig. 151. 


Fig. 152. 





























































































































PRACTICAL SOLUTIONS 


207 


waved in length, due to crooked growth, and the board 
passing through them in a plane, their edges crop out 
on the surface in irregular elliptic shape. 

Double Margined Doors are wide single doors framed 
to appear as pairs of doors. They are used in openings 
too wide proportionately for a single door, but where 
half the opening would be rather small for convenient 
passage. Figs. 150 to 154 show elevation and sections 
of a Double Margined Entrance Door, typical of the 
style in vogue in the latter part of the eighteenth and 
early part of the nineteenth centuries. 

These doors are made in two ways. In the earlier 
method the central imitation stile, which in this case 
is really a munting, is made in one piece and forked 
over the top and bottom rails, which are continuous. 
The intermediate rails are stubtenoned to the central, 
and through tenoned and wedged to the outside stiles, 
but unless the stub tenons are fox-wedged, the shoul¬ 
ders are very liable to start, for which reason the 
method of construction now to be described is generally 
preferred. The door is composed of two separate 
pieces of framing, each complete with two stiles and a 
set of rails that are tenoned through and wedged up. 
The two portions are then united by a ploughed and 
tongued and glued joint, which is hidden by a sunk 
bead in the centre, as shown in the diagram, Fig. 155, 
and the parts keyed together with three pairs of hard¬ 
wood folding wedges. The door is sometimes further 
strengthened by having flat iron bars sunk and screwed 
into the top and bottom edges. The actual process of 
putting the door together i's as follows: After the vari¬ 
ous rails and panels have been duly fitted and marked, 


268 


MODERN CARPENTRY 


each leaf is taken separately and the stiles knocked on. 
The one intended for the meeting stile having been 




Fig. 153. 



Fig. 155. Fig. 1.56. 


glued, is cramped up and wedged. Then the meeting 
stiles are made to a width, grooved, jointed, and re- 













































PRACTICAL SOLUTIONS 


269 


bated for the beads, as shown in the detail, Fig. 150. 
The ends of the tenons and wedges should be cut back 
Ys in. to prevent them breaking the joint when the stile 
shrinks. The mortises for the keys will have been 
made when the mortises for the rails were done, and 
cross-tongues are next glued in, the joint rubbed, the 
two stiles pinched together with handscrews, and the 
oak keys, well glued, driven in. At this stage the 
frames are stood aside to dry, after which the project¬ 
ing ends of the keys are cut off, the panels inserted, and 
the two outside stiles glued and wedged in the usual 
manner. After the door is cleaned off, the grooves to 
receive the beads are brought to their exact size with 
side rabbit and router planes.. Should iron bars be 
used, they are inserted in grooves made after the door 
has been shot to size. The bars should be about % in* 
shorter than the width of the door, so that the ends 
are not visible. 

Diminished stiles are sometimes cut out in pairs from 
a board or plank. When this is done, the back or out¬ 
side edge is shot straight and the setting out made 
thereon, the two portions being gauged to width also 
from the back, but the method more suitable for ma¬ 
chine working. Here the stile may be cut parallel to 
the full width, the face edge shot in the usual way and 
the setting out made upon that, the diminish being 
gauged from inside. In this method the mortises are 
made before the diminished part is cut out, to render 
that operation easier for the machinist. He should not, 
however, mortise right up to the sight lines on the di¬ 
minished part, because if the chisel is at all out of up¬ 
right when the waste is cut away, the mortise will be 


270 MODERN CARPENTRY 

found beyond the sight line, which will be a serious 
defect. 

To Set Out the bevelled shoulders of the stile and 
rail. Taking the stile first, having as described in the 
first method gauged and faced up the inside edges, and 
set out the width of the rails and mortises on the back 
edge, square over on each side the sight lines of the 
middle rail as at a a, Fig. 156. Then draw a second 



Stile and Rail of a Diminished 
Stile Door. 

Method of Obtaining Shoulder Lines. 

Fig. 156%. 

line representing the depth of sticking of the rail on 
the face side, and the rebate on the back side, as at b. 
Next run gauge lines down on each side of the dimin¬ 
ished part as working lines of rebate and sticking, and 
from the points of intersections of the stickings and 
rebates respectively, draw in the shoulder lines to the 
sight line of the lower edge of the rail at c. The only 
difference to be made when the stile is prepared by 
























PRACTICAL SOLUTIONS 


271 


the second method is that the sticking and rebate 
gauges would he run from the original face edge in¬ 
stead of the actual diminished edge, and as before 
stated, the sight lines would be marked on the face 
edge instead of the back, and squared down to the in¬ 
tersections. 

To Set Out the Rail. Mark on the bottom edge the 
“width” or sight lines of the widest part of the stiles. 
Square this point on to the upper edge as shown by 
dotted line in the sketch, Fig. 156, and set off there¬ 
from the amount of the diminish on the stile, as shown 
by the dotted line on the stile in the example: this is 2 
in. This line, knifed in on the edge, is the “sight line” 
of the upper part of the stile. Again set off beyond this 
line the amount of the sticking and rebate shown in 
the sketch by the lines e and f. Next run the sticking 
and rebate gauges on front and back sides, as shown at 
g, and square down the lines e and f to meet them. 
Then draw the shoulder lines from the intersections to 
the point d on each side. Having thus found the shoul¬ 
der line upon one rail, bevels may be set to them and 
used to mark any number. A contrivance sometimes 
used when a large number of similar shoulders have 
to be set out is shown in Fig. 157. This is known as a 
Shoulder Square. It consists of an ordinary set square 
provided with a movable fence or bar B, which is slot¬ 
ted to pass on both sides of the square, and is pivoted 
near the right angle. A set screw near the outer end 
of the bar passes through a concentric slot, and fixes 
the fence in any desired position. The pivot works 
tightly in a small slot to allow the lower edge of the 
bar to enter the right angle, and the outer edge of the 


272 


MODERN CARPENTRY 






































































PRACTICAL SOLUTIONS 


273 


square is also made a concentric curve to permit the 
easy passage of the end of the bar. The theory upon 
which the action of the tool is based is that the angle 
between the parallel bar B and either of the edges of 
the square is the complement of the remaining angle, 
the two combined forming a right angle which is the 



An Adjustable Square 
for Bevel Shoulders. 
Fig. 157%. 


desired angle between the edges of a rail and stile. Its 
application is shown in the upper part of the figure, a 
being the rail, b the stile. Either a rail or stile is first 
set out as described above, the edge of the square set 
to the shoulder line, and the bar brought up to the 
face of the work and fixed with the set screw when it 


























274 


MODERN CARPENTRY 


is ready for application to the other piece. The mould¬ 
ing upon a diminished stile should not be mitred but 
continued on to the shoulder, and the rail scribed over 
it, which will prevent an open joint occurring should 
the rail shrink. When doors, after knocking *together, 
are stored for a second season, a slight difference will 
have to be made in the setting out of the shoulders of 
the middle rail. The wider part of the stile will shrink 
more than the narrow part, and consequently if the 
shoulders are set out accurately at first as described 
above, when they are refitted the shoulders will be 
found short at the lower ends. To prevent this, allow 
about 1/32 in. extra on the lower part of the shoulder 
at each end of the rail. 

A Gothic Door of the Tudor period is shown in the 
elevation in Fig. 158 and section in Fig. 159. The head 
is four-centred. The upper panels are pierced tracery, 
and the lower ones carved drapery. The mouldings 
in this type of door are invariably stuck solid, and 
those on the stiles stopped at the sight lines of the 
rails. The mouldings on the latter are also frequently 
stopped at the muntings as shown especially in the 
earlier work. Many of the doors, however, of the Tu¬ 
dor period have the upper ends of the muntings mitred. 
In modern work of this style, when the mouldings are 
not stopped, it is usual to scribe them at the intersec¬ 
tion. Chamfers, however, are always stopped to ob¬ 
tain a square-built shoulder for the munting, as a 
shoulder scribed over a chamfer soon gets faulty 
through the shrinkage. 

Mediaeval doors were always constructed of oak,, 
but pitch or Georgia pine is now much used in this 



style of work. The older examples are mortised, ten¬ 
oned, and pinned together, wedging and glueing be¬ 
ing a modern invention. The joints at the head are 


































































































modern carpentry 


276 


usually slip tenons, pinned, or with dovetail keys in¬ 
serted. These joints in a modern door would be se¬ 
cured either with a hammer-headed key or handrail 
bolts. 



Fig. 160 shows the elevation of a superior five-pan¬ 
elled interior door with its finishings. Fig. 161 shows 
a vertical section through the opening, and Fig. 162 is 









































































































PRACTICAL SOLUTIONS 


277 



a plan showing in outline the framed soffit lining. Fig. 
163 is an enlarged section of one stile and part of 
panel, mouldings, etc. 


Fig. 163. Fig. 162. 



















































































































278 


MODERN CARPENTRY 


Revolving Doors.—An arrangement of vestibule 
doors, suitable for banks, hotels, etc., is shown in plan 
in Fig. 164. The doors are arranged at right angles to 
each other, and revolve around a vertical axis like a 
turn-stile. Curved side frames, each a little wider than 
a quarter of a circle, are fixed on each side of the door¬ 
way. A suitable width for the doors is 3' 6". The ad- 



Plans of Revolving Vestibule Doors. 

Fi &- 164 - Fig. 165. 

vantages of such an arrangement is that it is noise¬ 
less and draughtproof, the latter feature being obtained 
by having an india-rubber tongue fixed in the outer 
edge of each door. The doors are so hung that alter¬ 
nate doors can be folded back against the adjacent 
ones (Fig. 165), and thus give an uninterrupted pas¬ 
sage when required. 

Other Panelled Framing.—Framework filled in en¬ 
tirely with wooden panels, or with wooden panels in 
the lower part and glass in the upper part, is also re- 
quiied in the fittings for offices, for school partitions, 













PRACTICAL SOLUTIONS 


279 


and for screens in churches, business premises, etc. The 
arrangement of the framing is similar to that of doors, 
and the same terms are used to describe the various 
parts, the only difference being the proportions of 
height and width; these are, of course, governed by 
special requirements. 

Superior Doors.—In superior work, where the doors 
and surrounding framework are made of ornamental 
hardwood, it is often necessary to construct a door 
which shall be of one kind of wood on one side of the 
door and an entirely different kind on the other side. 
This would be necessary, for example, with a door 
opening from an entrance hall fitted entirely with oak 
into a room, the fittings of which must all be of walnut 
or mahogany. Such a door may be constructed in two 
thicknesses, each of the respective kind of wood, and 
each of a thickness equal to one-half of that of the 
finished door. The two parts are then secured together 
by tapering dovetailed keys, and the edges of the door 
are afterwards veneered to match the side of the door 
to which they correspond. Figs. 166 and 168 give de¬ 
tails of this kind of door. 

Grounds.—The architraves surrounding an opening 
are nailed to the lining, or where possible to the frame. 
In the best class of work, however, it is usual not to 
fix the door frames until the plastering is finished. 
Rough wooden battens or Grounds, of thickness equal 
to that of the plaster, are fixed to the walls around all 
door and window openings. These serve as a guide 
to the plasterer, and the door frames and the sur¬ 
rounding architraves are secured to them. When it 
it not desirable to have any nail holes visible in the 


280 


MODERN CARPENTRY 



Fig. 168. 











































































































































PRACTICAL SOLUTIONS 281 

finished surfaces, the door frames and architraves are 
fixed by screws. 

The fixing of the architraves around such a door¬ 
way affords a good example of Fixing by Secret Screw¬ 
ing. The mitres of the architraves are first glued and 
secured with dovetail keys or slip feathers. Stout 
screws are turned into the grounds about 12" apart, 
being left so that the head of the screw projects about 
half-an-inch in front of'the surface. On the back side 



of the architrave, exactly opposite the screw heads, 
small holes—equal to the size of the shanks of the 
screws—are bored; and about three-quarters of an 
inch below these, larger holes—of size equal to the 
heads of the screws—are bored. Each small hole is 
connected to the large one adjacent to it by a slot, the 
depth of which is slightly greater than the projec¬ 
tion of the screws. The architrave is fixed by placing 












282 


MODERN CARPENTRY 


it against the wall with the larger holes fitting on the 
screws, and then carefully driving it down so that the 
heads of the screws hook into the fibres behind the 




Vertical Section 

Fig. 171. 


slots. By placing the screws so that they are slightly 
inclined, the tendency is to draw the architraves closer 
to the wall. Fig. 169 shows the explanatory detail. 

The above remarks upon door frames, linings, etc., 
apply especially to the doors of dwelling-houses. Door 
























































































PRACTICAL SOLUTIONS 


283 


frames for warehouses, workshops, outbuildings, etc., 
do not as a rule require linings or architraves, a small 
fillet being nailed into the angle between the door 
frame and the wall instead. Vestibule doors are often 
hung to swing both ways, and the door frames have 
a hollow rebate or groove in the middle of the width 
of the frame, to receive the rounded edge of the door 
(Figs. 170, 171 and 172). Many of the heavier kinds 
of framed and ledged doors are not provided with 
wooden frames but are hung with bands and gudgeons, 
or arranged to run on pulleys as described elsewhere. 



Details of a pair of Vestibule Doors with Side-lights. 
Fig. 172. 


Fig. 173 shows an ordinary sash door with three 
panels below. Fig. 174 shows a section of door and 
frame. Fig. 175 shows plan of door and part of cross 
section of frame. Fig. 176 shows the height and width 
rod of a door, which guides the workman in laying out 
his work. 

External Doors are invariably hung in Solid Frames. 
Internal doors, chiefly to build up casings or Linings, 
of comparatively thin substance. In certain positions, 
such as vestibules and shop-fronts, where there are no 
wall openings to line, solid frames are also used for 





284 


MODERN CARPENTRY 





Fig. 173. 


























































































































































PRACTICAL SOLUTIONS 


285 



Fig. 174. 


Height and Width Rod of a Door. 
Fig. 176. 
















































286 


MODERN CARPENTRY 


interior doors. The members of solid frames are usu¬ 
ally made of square section or slightly thicker than 
wide; this arrangement may, however, be varied to 
meet the necessities of the design; the rebates, stops, 
mouldings, &c., are worked in the solid. The outer 
vertical members of these frames are called josts or 
jambs; interior ones, mullions. The horizontal mem¬ 
bers are sill, transom and head. The jambs are framed 
between the head and sill, chiefly that the ends or 
horns of the latter may run beyond the frame, and so 
provide fixings that can be built into the wall; and 



Fig. 177. 

also because the shoulders of the post form a better 
abutment for carrying any load that may be thrown 
on the head than the edge of a tenon would. Transoms 
are cut between the jambs and also between the mul- 
lions when these are used. 

A Segment-headed Frame is shown in Fig. 177, and 
enlarged details of the joints in Figs. 178, 179, 180. The 
heads of these frames are cut out of the solid when 
the rise will permit of their being cut from pine of 
ordinary width. When this cannot be done, they are 

















PRACTICAL SOLUTIONS 


287 


made in two lengths, jointed at the crown, and fas¬ 
tened with a handrail bolt. The horns are taken out 
level at the springing line, and the back is made 
roughly parallel to the shoulders for convenience in 
fixing. When the frame is 4 in. and upwards in thick¬ 
ness, Tlouble tenons should be used, as shown in Fig. 
179; and if the position in which the frame is to be 
fixed does not admit of the horns being left on, the 
mortises should be haunched back, as shown in Fig. 



180, although the horns would not be cut off until the 
frame was fixed, as they would be required for the 
purpose of cramping up the frame. 

A Semi-headed Frame may have its head cut in two 
or three lengths and bolted together, but is frequently 
built up as shown in Figs. 181, 182, 183. The jambs 
and transom are worked solid, but the head is formed 
in two thicknesses glued and screwed together, one 
























288 


MODERN CARPENTRY 


layer being in two lengths, the other in three, so as 
to break joint. This is both a strong and economical 
way of forming a head, because the grain is less cut 
across than it would be in a head cut out of one thick¬ 
ness, and the labor of rebating is also dispensed with, 
the inner ring being kept back % in. to form a re¬ 
bate. The head is fastened to the jambs by hammer¬ 
head tenons.and Shoulder tongues, as shown in Fig. 
182, and double tenons are used for the transom to 



A Semi-headed Solid Frame. 

Fig. 181. Fig. 182. 



Fig. 183. 


•avoid cutting the root of the head tenons. The tran¬ 
som is better kept about 3 in. below the springing as 
shown, to ensure a strong joint. But if the exigencies 
of the design necessitate its being placed at the spring¬ 
ing, then the jamb should be carried above the spring¬ 
ing and a portion of the curve worked upon it, because 





























PRACTICAL SOLUTIONS 28$ 

if the two joints are made together, the connection will 
be very weak. 

There are four varieties of door casings or Jamb Lin¬ 
ings as they are also termed, viz., Plain, Framed, 
Double Framed, and Skeleton Framed, these names de¬ 
fining the method of construction. Other sub-names are 
also used denoting the nature of the ornamentation, as 
in doors, with which they agree in the general arrange¬ 
ment of their parts. The term Plain is applied to any 
wall lining however it may be treated, if it is made 
of one flat board or surface. 

A “Set”of Linings comprise a pair of jambs and a 
head or soffit lining. The flat, against the edge of 
which the door rests when closed, is called the Stop, 
and in common work these are merely nailed upon the 
surface of the main lining, being kept back from the 
edge sufficiently far to form a rebate for the door. In 
better work the rebate is worked in the solid, the lin¬ 
ing in such case being thicker. Not less than iy 2 in. 
stuff should be used for any lining to which a door 
has to be hung, as the rebate takes y 2 in. out of the 
thickness, leaving only 1 in. for screw hold for the 
hinges. This, however, may be supplemented by hinge 
blocks glued to the back of the lining just behind 
where the hinges will be inserted, as shown on one side 
of the isometric sketch of a set of Plain Jamb Linings 
(Fig. 184). When the lining is rebated on both edges, 
it is said to be “double rebated.” Plain linings are 
not suitable for walls thicker than 14 in., in conse¬ 
quence of the amount of shrinkage which disarranges 
the finishings, and even in 14 in. work they are better 
framed. 



290 MODERN CARPENTRY 



Rebated Set of Plain Linings with Double 
Set of Grounds. 


Fig. 184. 































PRACTICAL SOLUTIONS 


291 


Grounds.—Figs. 188 and 189 are light frames form¬ 
ing a boundary to all openings in plastered walls, their 



Fig. 185. 

purpose being to act, as the name implies, as a ground¬ 
work for the linings, architraves, &c., and also as a 



Fig. 186. 

margin and gauge for the thickness of the plaster. 
They are either bevelled or grooved on the back edges 
to form a key for the plaster, and should be framed up 











































292 


MODERN CARPENTRY 


perfectly square, and with faces and inside edges true, 
straight and square, to ensure good fittings in the fin¬ 
ishings. They are fixed by nailing to plugs or wood 



End of a Jamb Lining showing Tongues. 
Fig. 187. 


bricks in the walls, and should be secure, plumb and 
out of winding. They are usually prepared out of 1 in. 



Elevation and Edge View of a Set 
of Grounds. 

Fig. 188. Fig. 189. 


stuff, their width depending upon the width of the 
architraves, which should overhang them about % in. 
More cover than this is not advisable, or the fixing for 












































PRACTICAL SOLUTIONS 293 

the outside of the architrave will be lost, and for the 
same reason much bevel should not be given to the 
back edge. A *4 in. will provide quite sufficient key 
for the plaster, and also, as the grounds have to be 
bevelled before glueing up, a very thin edge is liable 
to be crushed in cramping them up. The sets of 
grounds on each side of an opening are connected 
across the jamb by 1 and 2 in. slips called Backing 
pieces, dovetailed, and nailed to the edges. Grounds 



are, however, very often fixed flush with the brickwork, 
which does not permit of the use of backings; it is a 
very inferior method, the linings either touching the 
walls, or if made smaller, having to be fixed with fold¬ 
ing wedges and firring pieces, which do not afford so 
firm a fixing. When the architrave is of such width 
that the sides of the ground are required more than 
5 in. wide, it is not advisable to make them in one piece, 
because from their position they are very liable to cast, 
and so spoil the appearance of the finishings. In such 
cases, the grounds should be formed as skeleton frames 


































294 


MODERN CARPENTRY 


with rails and stiles, where required from 3 to 5 in. 
wide (see Fig. 190). 

Grounds are sometimes moulded or beaded on their 
inside edges, as in Fig. 191. They are then called 
Architrave Grounds or Moulded Grounds, and are fixed 
to the edges of the linings, which in such cases must 



Fig. 191. 


Fig. 192. 


be fixed first. The moulded ground is more frequently 
used m window and shutter openings than in door 
openings. All joinery finishings are fixed to grounds, 
but these are 'seldom framed, except in the cases of 
door, window and shutter openings, all other work 
being fixed to rough grounds, which will be illustrated 
in their appropriate place. 
























PRACTICAL SOLUTIONS 


295 


Framed Grounds are usually wedged up and sent 
out from the shop in pairs, as shown in Pig. 192, Fig. 
193 being an edge view to larger scale. Two sets are 
lightly nailed face to face, and with heads reversed, 
in which position they can be knocked apart and glued, 
then cramped, squared, and wedged up with facility. 

Architraves are the moulded borders or frames to 
window or door openings. When square at the back, 
as in Fig. 191, they are termed Single-faced; when 
moulded at the back, as in Fig. 185, Double-faced. No 



Framed Architrave 
and Grounds. 


Fig. 193. 


architrave should be made wider than 6 in. in one piece, 
as there is great danger of its splitting when shrink¬ 
ing, being fixed necessarily at both edges; but when 
this size is exceeded, the moulding should be made up 
of two or more members, grooved and tongued to¬ 
gether at some convenient point, as shown in Figs. 185 
and 193, the latter being an example of a very wide so- 
called Framed Architrave;' then, if fixed at both edges, 
it will be free to shrink in the middle. 













296 


MODERN CARPENTRY 


Heavy Architraves are ploughed and cross tongued 
in the mitres, and in some cases framed or stub mortised 
and tenoned, as shown in Figs. 194 and 195. These 
are glued and fixed with screws turned in from the 
back. Small handrail bolts are also used to draw up 
the mitres in thick mouldings. Sets of architraves so 
treated are framed up on the bench, provided with 
stretchers at the bottom end, and braced to prevent 



An Architrave Tenoned and 
Mitred. 

Fig. 194. Fig. 195. 


racking during transit, and are fixed complete. Smaller 
ones are mitred and fixed piece by piece. In the best 
class of polished work, secret fixings are used for se¬ 
curing the architrave to the grounds. These are shown 
in Figs. 196, 197 and 198. A number of stout screws 
are turned into the grounds with their heads project¬ 
ing uniformly about % in. Corresponding holes and 
bevelled slots are made in the back of the moulding, 
which is dropped on the screws, and carefully driven 





























PRACTICAL SOLUTIONS 


297 


down in the direction of the arrow. The architrave 
should be driven on to the screws dry in the first place; 
two workmen being employed in the operation, one 
upon a scaffold carefully drawing down the architrave 



Elevation of Back of Architrave. —> Section of 
Architrave, and Elevation of Face of Grounds. 

Fig. 198. Fig. 197. Fig. 196. 


equally on each side; the other pressing the moulding 
tightly towards the grounds, and striking a piece of 
wood with a heavy hammer over each screw as his fel¬ 
low strikes the top. After being duly fitted the archi¬ 
trave may be knocked up again, and the front edges 
only lightly glued before replacing. 

Foot or Plinth Blocks are used at the bottom of all 
architraves in good work, and are secured in two ways 

























298 


MODERN CARPENTRY 


(see Figs. 199 and 200). The latter is the better way, 
as there is less difficulty in fitting the shoulder, and 
also less danger of splitting the plinth, when driving 
the dovetail tenon in; this is glued and secured with 
a screw turned in the back. A handscrew may be 
used to grip the sides of the block whilst driving the 
latter on to prevent its splitting. 



Fig. 199. Fig. 200. 


Door and Jambs Complete. Figs. 201 and 202 show 

a door, with jambs and linings complete. The details 
for constructing and fastening in place are all shown. 
Each member of the door is named, also base and 
grounds. 

Figs. 202 to 220 show a number of typical doors, some 
of which will be found suitable for any particular pur¬ 
pose. 






























PRACTICAL SOLUTIONS 299 



Fig. 201, 


Fig. 202. 




























































































































































































300 


MODERN CARPENTRY 



Fig. 203. 



Fig. 204. 



Fig. 205. 
































































































































PRACTICAL SOLUTIONS 


301 






































































































302 


MODERN CARPENTRY 



Fig. 218. 


Fig. 219. 


Fig. 220, 



























































































































PRACTICAL SOLUTIONS 


303 



\ 








































PART III. 


MECHANICS OF CARPENTRY. 

In this part I intend to give the reader some prac¬ 
tical instructions in “ The ‘ Mechanics of Carpentry/’ 
with the methods of calculating the strength of trusses, 
and timbers for structural purposes. All the rules 
shown have been taken from the best authorities, as 
Tredgold, Barlowe, Hurst, Riley, Trautwine, Hat£eld, 
Kidder, Nicholson and others. I have, in a few in¬ 
stances, corrected, and added to some of the propo¬ 
sitions, which I have found in practice, to be neces¬ 
sary. 


INTRODUCTION TO THE MECHANICS OF CAR¬ 
PENTRY. 

It is well known that some members of a framed 
structure must be made stronger than others. The 
reason is that the weights or other forces acting on a 
truss differ from each other in magnitude and direc¬ 
tion. It is obviously necessary, therefore, to be able 
to estimate the various forces acting, so that the mem¬ 
bers may be made of the required strength without 
undue waste of material. The general principles un¬ 
derlying the measurement of forces may with advan¬ 
tage now be considered briefly. 

305 


306 


MODERN CARPENTRY 


The Nature of Force. Force may be defined as that 
which moves, or tends to move, a body at rest, or 
which changes, or tends to change, the direction or 
rate of motion of a body already moving. A familiar 
example of force is met with in gravitation, whereby 
an object has a tendency to fall to the ground. In 
order to support it, an upward force equal to the 
weight of the object must be exerted. The phrase 
“equal force” implies that forces can be measured. 
In this country they are usually measured in terms of 
weights in lbs., cwts., etc. Any one who has seen a 
pulley, or lever, at work knows that the direction of 
application of a force can be changed. Evidently, 
then, forces can be represented graphically. Lines 
drawn to scale are employed and these can be ar¬ 
ranged to exhibit at the same time both the magnitude 
and the direction of the forces. Thus, a weight of 10 
lbs. acting vertically downwards can be represented 
by a vertical straight line 10 units in length If the 
unit of length be the line will measure ten times 
\ whereas, if the unit of force be represented 
by a length of 1", the graphic representation of the 
force will be a vertical straight line 10" long. 

Resultant of two or more Forees. (1) When two or 
more forces together act at a point in the same direc¬ 
tion and in the same straight line, the resultant force 
is equal to the sum of the components. 

Example, (a) If two 10 lb. weights attached to a 
cord are hung upon the same nail, the resultant weight 
acting upon the nail is 10+10=20 lbs. 

(2) If two equal forces together act at the same 
point in opposite directions, but in the same straight 


MECHANICS OF CARPENTRY 


307 


line, they neutralize each other, and the forces are said 

to be in equilibrium. 

Example. A spring balance carries a weight of 6 
lbs. The index finger of the balance shows that the 
spring exerts an upward force equal to the downward 
force—the weight; and a state of equilibrium is ob¬ 
tained. 

If the two unequal forces together act at the same 
point in opposite directions, but in the same straight 
line, the resultant force is equal to the difference.be¬ 
tween the forces, and is in the direction of the greater. 

It is evident, then, that the directions of the forces, 
and therefore the angles they make with one another, 
must be considered in determining the forces acting at 
any given point. 

If a flexible string be attached to a weight, and then 
passed over a frictionless pulley, there will be the 
same tension in every part of the string, irrespective 
of any change of direction caused by using the pulley. 

To illustrate these facts clearly, suppose that two 7 
lb. weights, connected by a cord, hang over a smooth 
peg as shown in Fig. 1. The total weight on the peg, 
neglecting the weight of the cord (which may thus 
be any length), is 14 lbs., the sum of the two weights. 

Again, suppose three such pegs in a horizontal 
straight line, and the cord and weights to be passed 
over them as shown in Fig. 2. Evidently the weight 
on the central peg is nothing. Now, suppose the out¬ 
side pegs to be lowered slightly, as shown by dotted 
lines in the figure; the central peg will now carry a 
small proportion of the weight, and the more the out¬ 
side pegs are lowered, the niore weight will be thrown 


308 


MODERN CARPENTRY 


on the central peg, until, as shown in Fig. 1, it carries 
all the weight, i. e., 14 lbs. Therefore the weight upon 
the central peg varies according to the direction of the 
forces acting on it—from nothing in Fig. 2 to 14 lbs. in 
Fig. 1. 

The magnitude and direction of the resultant force 
acting upon the central peg, and upon each of the out¬ 
side pegs, can be determined by the parallelogram of 
forces. 



The Parallelogram of Forces.—If two forces acting 
at a point be represented in magnitude and direction 
by the adjacent sides of a parallelogram, the resultant 
of these two forces will be represented in magnitude 
and direction by that diagonal of the parallelogram 
which passes through the point. 























MECHANICS OF CARPENTRY 


309 


Example 1. The angle at A, when the cord passes 
over the pegs B 3 , A,C 3 , shown by the dotted lines in 
Fig. 2, is given. Determine by the parallelogram of 
forces the stress on the peg A, i. e., the single force 
acting through the point A, which shall be equal in 
effect to the forces AB 3 , AC 3 acting together. 

Produce AB 3 and AC 3 , and mark off on each line 7 
units, measuring from A. Then A1 and A2 represent 
in magnitude and direction the forces caused by the 
loads. Complete the parallelogram by drawing ID 
parallel to A2, and 2D parallel to Al. The length of 
the diagonal AD, measured in the same units as the 
lines Al and A2, represents the magnitude of the re¬ 
sultant force—i. e., the stress on the peg A. The di¬ 
rection of the force will obviously be downwards. A 
force represented in magnitude and direction by D A 
would evidently counterbalance the force A D, and 
would therefore counterbalance Al and A2 acting to¬ 
gether. Forces which balance each other are said to 
be in equilibrium. 

Example 2. Determine the magnitude and direction 
of the single force which will replace the two forces 
exerted by the cord and weight on the peg B 3 (Fig. 2). 

Draw ab 7 units long (Fig. 3) and parallel to the 
cord Al in Fig. 2. From b draw be also 7 units long 
and parallel to the cord below the peg B r Complete 
the parallelogram by drawing dc and ad parallel to 
ab and be respectively. Then the diagonal bd gives 
the magnitude of the required force, the direction of 
which is from b to d. 


,310 


MODERN CARPENTRY 


Example 3. Fig. 4 shows the application of the 
parallelogram of forces to determine the resultant 
force on the peg B. 

In the above examples no allowance has been made 
for the weight of the cord or for the friction on the 
pegs. It is assumed in each case that the forces are 
acting at the point of intersection of the straight lines 
produced. 




Example 4. Two forces of 10 and 6 lbs. respectively 
act from a point and in directions which are at right 
angles to each other. Determine the magnitude and 
direction of the single force which can replace the two 
forces. 

Let the line AB (Fig. 5) represent in magnitude a 
force of 10 lbs. acting at the point A in the direction 
indicated by the arrow, and A C a force of 6 lbs. acting 





MECHANICS OF CARPENTRY 


311 


at right angles to AB. Complete the parallelogram 
ACDB. Then the length of the diagonal AD rep¬ 
resents the magnitude of the resultant force, and the 
direction in which it acts will be from the point A, as 
shown by the arrow. 

It must be understood clearly that a resultant is a 
force which can take the place of, and will produce 
the same effect as, two or more forces. To maintain 
equilibrium, the resultant force must be counterbal¬ 
anced by an equal force acting in the opposite direc¬ 
tion. The force so acting is called the equilibrant. 



Example 5. Figs. 6 and 7 show the magnitude and 
direction of the resultant force when forces of 9 and 6 
lbs. respectively act at angles of (a) 120°, (b) 45°. 

The simple apparatus shown in Fig. 8 clearly illus¬ 
trates the principle of the parallelogram of forces. On 
a vertical board are fixed two small pulleys by means 
of screws, so that they revolve with as little friction as 
possible. By making a ^three-way string, passing it 
over the two pulleys, and adding varying weights to 



312 


MODERN CARPENTRY 


each of the three ends of the string, it can be demon¬ 
strated clearly how the three forces act. In Fig. 8 the 
weights are respectively 5, 6, and 4 lbs. By drawing 
the parallelogram A B D C, such that A B equals 5 
units in length, and A C equals 4 units, the diagonal 
D A is found to measure 6 units, and to represent the 
magnitude of the middle weight. If other weights are 
attached to the ends of the strings, different results 
will, of course, be obtained. 



Triangle of Forces. The triangle of forces is used to 
determine the magnitude and direction of any three 
forces which balance each other. The rule may be 
stated as follows: If three forces acting at a point are 
in equilibrium they can be represented in magnitude 
and direction by the three sides of a triangle taken in 
order. 






MECHANICS OF CARPENTRY 


313 


Example 1. The forces acting upon A (Fig. 8) are 
in equilibrium. 

Since the length of the line A B=5 units, and the 
line B D is parallel and equal in length to A C=4 units, 
and the diagonal D A is in a line with the direction of 
the middle vertical weight and equal in length to 6 
units; then the sides A B, B D, and D A of the tri¬ 
angle A B D represent both in magnitude and direction 
the forces acting at the point A. 



To save confusion it is usual, however, to draw a 
separate triangle to illustrate these forces. A some¬ 
what different system of lettering also simplifies the 
consideration of the examples. This is known as Bow’s 
notation. In it the two letters denoting a force are 
placed on each side of the line representing the fore** 









314 


MODERN CARPENTRY 


that is, in the spaces between such lines. Thus in Fig. 
9 the three forces acting at the point o are referred to 
as A B, B C, C A respectively, i 
Example 2. Given the magnitude (9 lbs.) and the 
direction (indicated by the arrow) of A B, and the 
angles which the directions of the three forces make 
with each other, it is required to find the magnitude 
and direction of B C and C A when the forces are in 
equilibrium. 




Draw the line a b (Fig. 10) parallel to the direction 
of action of the force A B, 9 units long, and in the 
direction shown by the arrow. From b draw b c paral¬ 
lel to B C until it meets a c drawn parallel to C A. Then 
the triangle a b c is the triangle of forces, and the 
direction of the forces B C and C A can be found by 
taking the sides of the triangle in order, viz., a to b, 
b to c, c to a; and these directions give also the direc¬ 
tions of action of the forces represented by the lines 
parallel to a b, b c, and c a respectively. Thus A B acts 
* rota tlie d°int o; B C acts towards o; and C A acts 
from o. 






MECHANICS OP CARPENTRY 


315 


The following examples show the application of these 
principles to simple practical questions. 

Example 3. A rope bears a tensile stress (pull) of 
30 cwts. Find the magnitude of the stress in each of 
two ether ropes which make an angle of 60° with each 
other, and together balance the stress in the first rope, 
supposing the second and third ropes are equally 
stressed. 



Fig. 11. 


Fig. 11 shows the application of the triangle of forces 
to the solution of this question, the answer giving the 
stress in each rope as 17.32 cwts. By going round the 
sides of the triangle in order, it will be seen that the 
force in each of the three ropes acts from the joint. 

Example 4. A buckling^chain is used to raise heavy 
blocks of stone. What is the amount of stress in the 




316 


MODERN CARPENTRY 


links of the chain when raising a weight of one ton, if 
the buckling-chain is: 

(a) pulled tightly as in Fig. 12; 

(b) placed loosely round the stone as in Fig. 13. 



The correct solution of this question depends on (1) 
the weight of the stone; (2) the angle between the 
forces A C and B C. 



The application of the triangle of forces in each case 
(Figs. 12 and 13) shows that the stresses A C and B C 
are more than twice as great when the chain is fixed 














MECHANICS OF CARPENTRY 


317 


as in Fig. 12 as they are with the arrangement in Fig. 
13; or, the tighter the chain—i. e., the greater the angle 
between the forces B C and C A—the greater is the 
stress on the links. 

Example 5. A triangular bracket fixed against a 
wall, as shown in Fig. 16, has a weight of 5 cwts. sus¬ 
pended from the outer end o. What is the nature and 
amount of stress in each of the members o A and o B! 

Fig. 15 is the triangle of forces used to determine 
these stresses, and is drawn as follows: 1^ is drawn 




parallel to and represents the downward force (the 
weight of 5 cwts.) to scale. From 2 X draw 2j3 x parallel 
to 2 3 in Fig. 14 until it meets 1^ drawn parallel to 
1 3. Then the triangle 1 1 2 1 3 1 represents the magnitude 
of the forces. 

By going round the triangle in order as shown by the 
arrows, we find that 2 3 acts towards* the joint o and 
is therefore a compression stress or thrust, and 3 1 acts 
from the joint and is therefore a tension stress or pull. 








318 


MODERN CARPENTRY 


Fig. 16 shows a somewhat modified design of trian¬ 
gular wall-bracket, and Fig. 17 is the triangle of forces 
by which the stresses in the various members are ascer¬ 
tained. 




Fig. 17. 


Example 6. What is the nature and amount of stress 
in each of the members A B and A C (Fig. 18) caused 
by the weight of 10 cwts. acting as shown? 

This example may be taken as typifying a simple 
kind of roof-truss with the weight taking the place of 
the ridge piece. Re-letter or figure the diagram accord¬ 
ing to Bow’s notation. Draw the vertical line 2 1 3 1 , 
equal in magnitude and direction to the weight 2 3. 
Complete the triangle by drawing lines parallel to the 
members AC and AB, from the points 2 1 and 3 A re¬ 
spectively. These lines represent the amount of stress 
along the members A C and A B. On taking the sides 
of the triangle in order as shown by the arrows, it is 







MECHANICS OF CARPENTRY 


319 


seen that 2 1 3 1 act downwards; 3 1 1 1 acts towards the 
joint A, as does also 1^; therefore each member is 
subject to a compression stress (thrust). 

Fig. 19 shows another example of this kind with a 
much smaller angle between the forces. 

Fig. 20 illustrates a still further example, where the 
two sides are of unequal inclination. 



Polygon of Forces. The method of obtaining the 
resultant of any two forces acting at a point can be 
extended to three, four, or any number of forces. 

Example. 0 A, OB, 0 N C, 0D, 0E, (Fig. 21) repre¬ 
sent the magnitude and direction of five forces acting 









320 


MODERN CARPENTRY 


at the point O. Determine the magnitude and direction 
of the resultant force. 

This problem can be solved either by an application 
of the parallelogram of forces or by a direct construc¬ 
tion. 

(1) Determine by the parallelogram of forces, the 
resultant 0 1 of forces OA and OB (Fig. 22). Simi¬ 
larly, determine the resultant 0 2 of the forces 0 1 and 




0 C. Again, 0 3 is the resultant of the forces 0 2 and 0 
D; and finally 0 4 is the resultant of 0 3 and 0 E. 
Therefore, 0 4 is the resultant of all the original forces; 
or, in other words, a single force equal in magnitude 
and direction to the force 0 4 will have the same effect 
at the point 0 as the five forces have when acting to- 







MECHANICS OF CARPENTRY 


321 


gether. Since a force 4 0 will balance 0 4, a force rep¬ 
resented in magnitude and direction by tbe line 4 0 
will, together with the five given forces, produce equil¬ 
ibrium at the point 0. 

(2) The same result may be obtained more simply as 
follows. Re-letter the forces as shown in italics (Fig. 
J2), and then, as in Fig. 23, draw a straight line a'b' 



equal in magnitude and parallel to a b (Fig. 22). From 
b' draw b'c' equal and parallel to b c; continue the pro¬ 
cess, taking the forces in order. It will be found by 
drawing the closing line of the polygon, that is, by 
joining x' to a', that x'a' gives the magnitude and the 
direction of the force required to produce equilibrium. 
Conversely, a'x' is the resultant of all the original 
forces. By drawing the line 4 0 through the point 0 
(Fig. 22) and indexing it to scale, the required result- 




322 


MODERN CARPENTRY 


ant—which corresponds with the one determined by 
the parallelogram of forces—is obtained. Its direction 
is indicated by the arrow. 

Fig. 25 is the polygon of forces when two of the 
forces, be and d e, act towards the joint (Fig. 24), 
the magnitude of all the forces being as in the previous 
example. In this case the equilibrant is determined, 
and is shown by the thick line in Figs. 24 and 25. 




Figs. 21 and 25 should be compared carefully. 

The polygon of forces may be stated as follows: If 
two or more forces act at a point, then, if starting at 
any point a line be drawn to represent the magnitude 
and direction of the first force, and from the point thus 
obtained another line be drawn similarly to represent 
the second force, and so on until lines have been drawn 
representing each force,—the resultant of all these 
forces will be represented by a straight line drawn 
from the starting point to the point finally reached. 

Polygons, parallelograms, or triangles, of forces, 
when used to determine either the resultant or the 
equilibrant of stresses acting at a point, are called 
reciprocal diagrams. 




MECHANICS OF CARPENTRY 


323 


Inclined Forces in one Plane but not acting through 
one Point. The foregoing examples deal only with 
forces which act at a single given point, and in these 
cases the resultant acts at the same point. When all 
the forces do not act at the same point, the magnitude 
and direction of the resultant is obtained as in previous 
examples, i. e., by drawing the reciprocal diagram; the 
line of action, however, still remains to be determined. 
To determine this line of action, it is necessary to draw 
what is known as the funicular or link polygon. The 
method is as follows: 



Fig. 26. 


Fig. 27. 


Example. Let X, Y, Z (Fig. 26) be three forces in 
the same plane and of the magnitude and direction 
shown. It is required to find the magnitude and the 
line of action of the resultant force. 

Re-letter the forces abed according to Bow’s nota¬ 
tion and draw the reciprocal diagram a'b'c'd'; the line 






324 


MODERN CARPENTRY 


a'd' which closes the figure represents the magnitude 
of the resultant. To obtain the actual line of action 
of the resultant, take any point or pole O and join 
a'O, b'O, c'O, d'O. The figure thus obtained is called 
the polar diagram. The funicular polygon is now con¬ 
structed by drawing—anywhere in the space b—a line 
1 2 parallel to b'O and intersecting the forces N and Y 
at 1 and 2 respectively. From 2 draw 2 3 parallel to 
c'O, intersecting the force Z in 3. Through 1 draw 
1 4 parallel to a'O, and through 3 draw 3 4 parallel to 
d'O. Through the point of intersection 4, draw a line 
R parallel to a'd'. R is the required line of action of 
the resultant of the three given forces, and its magni¬ 
tude is represented by the length of a'd'. 



Fig. 28. 


Parallel Forces. In addition to forces acting in the 
ways already explained, it is necessary to consider a 
few examples of parallel forces. (These- must not be 
mistaken for those dealt with by the parallelogram of 
forces, as they are entirely different.) 

In all the examples of parallel forces now to be con¬ 
sidered, the forces will act vertically. As these can be 
shown easily both graphically and arithmetically, each 
example will be worked out by both methods. 

The simplest examples of the equilibrium of parallel 
forces are found in the use of levers. The lever shown 
in Fig. 28 is a straight bar resting on a triangular 




mechanics of carpentry 


325 


block, F, called a fulcrum. At the ends A and B of the 
lever, forces P and W respectively act vertically down¬ 
wards. It is plain that forces P and W will tend to 
rotate the lever in opposite directions around the fixed 
point F. The tendency of either force to rotate the 
lever is called the moment of that force; it is measured 
by the product of the force into the perpendicular dis¬ 
tance (called the arm of the force) of the fixed point 
from the line of action of the force. When the two 
moments are equal the lever is in equilibrium. The 
conditions of equilibrium therefore are: 

pxaf=wxbf. 


If A F be 6", B F be 2", and W=9 lbs., then P will re¬ 


quire to be 


9X2 
6 ‘ 


3 lbs.: the moment of each force 


being 18 inch-lbs. 



B 



4lb $. 


Fig. 29. 


Since moments are always expressed in terms of the 
product of a force and a length, both these factors 
enter into every statement of the magnitude of a mo¬ 
ment. If the distances be expressed in feet, and the 
forces in cwts., the moments will, of course, be ex¬ 
pressed in ft.-cwts., and so on. 

Example 1 . A horizontal bar 3 ft. long has a weight 
of 2 lbs. at one end, and of 4 lbs. at the other end (Fig. 
29). Find the point at which the bar must be sup- 







326 


MODERN CARPENTRY 


ported so that it will rest horizontally. (Neglect the 
weight of the bar.) 

Arithmetically.—Since the lever is in equilibrium, 
the total downward force of 6 lbs. is balanced by an 
upward force (reaction) of 6 lbs. at the unknown point 
of support, and the moment of the upward reaction, 
about any point, is equal to the sum of the moments 
of the downward forces about the same point. Con¬ 
sider moments about A: 

Moment of weight at A, about A,=2X0. 

" “ “ B “ A,=4XAB. 

“ reaction about A=(2+4)XAX. 

6XAX=(2X0) + (4XAB); 
6AX=0+(4X3); 

AX=12/6=2 feet 

Graphically.—In the consideration of these forces 
graphically, fhe polygon of forces becomes a straight 
line. A polar diagram and a funicular polygon are re¬ 
quired. 

Fig. 30 shows the bars with the weights suspended. 
Letter the forces A B and B C. The polar diagram is 
drawn as follows: Draw to a suitable scale, a vertical 
line a c, 6 units long—equal to the sum of the weights. 
From any point 0 which may be at any convenient dis¬ 
tance from a c, draw 0 a, Ob, and 0 c. To construct 
the funicular polygon, draw, in the space B, 1 2 parallel 
to bo. From 1 draw 13 parallel to a 0, and from 2 
draw 2 3 parallel to 0 c, and produce it to intersect 
1 3 in 3. Then the vertical line drawn through the 
point 3 will give the position of the fulcrum. If the 
distances from this point to the points of application of 


MECHANICS OF CARPENTRY 


327 


the weights be measured, it will be found that they are 
in inverse proportion to the magnitudes of the 



weights, and that the weight on the right hand side of 
the fulcrum, multiplied by its arm of leverage, will be 
equal to the weight on the left hand side, multiplied by 
the arm of leverage on that side—the arms being one 
and two feet respectively. 

A _ B r h 

'l 

albs. 3 Ids gibs. 4-lbs. 

Fig. 31. 

Example 2. Four weights of 2, 3, 6, and 4 lbs. re¬ 
spectively hang on a bar as shown in Fig. 31. Deter¬ 
mine the point at which the bar must be supported to 
rest horizontally, the weight of the bar being neglected. 














328 


MODERN CARPENTRY 


Arithmetically.—Let the required point of support 
be denoted by the letter X. When the bar is in equil¬ 
ibrium, the sum of the moments about A of the down¬ 
ward forces, must be equal to the moment, about A, of 
the upward reaction at the point of support. 

the downward moments about A 
= (2X0) + (3X7) + (6X11) + (4X15) 

= 0 + 21 + 66 + 60 =147. 

The moment about A of the upward reaction 
=Sum of all the weights XAX 
= (2+3+6+4) X AX=15AX; 

15AX=147; 

:. AX=147/15=9 4/5 feet. 



Graphically.—Draw the vertical line of loads e a, 
representing to scale the sum of the weights as shown 
(Fig. 32). Construct the polar diagram by drawing 







MECHANICS OF CARPENTRY 


329 


from any point 0 the lines e 0, d 0, c 0, b 0, and a 0. 
To draw the funicular polygon, draw vertical lines un¬ 
der each weight, and—starting anywhere in the line of 
the first weight as at 1—draw in the space B a line 1 2 
parallel to 0 b; in the space C draw 2 3 parallel to 0 c; 
in the space D draw 3 4 parallel to 0 d. Through 4 
draw 4 5 parallel to e 0 and through 1 draw 1 5 parallel 
to 0 a. The vertical line drawn through the point 5, 
where these two lines meet, gives the position of the 
point of support. 

Although the application of the lever as a tool or 
machine is an everyday occurrence with the workman 
in such appliances as the turning bar of the bench-vice 
and sash-cramp, the screw-driver, brace, pincers, claw- 


JC 


W=/OCwts 




6ft 


Fig. 33. 


hammer, grindstone, treadle lathe, mortising-machine, 
etc., the detailed consideration of each of these cannot 
be entered into for want of space. The following 
examples involving the use of the crowbar will suffice 
further to illustrate the principles involved. 

Example 1. What force must be exerted at one end 
of a crowbar 6 ft. long, to raise a weight of 10 cwts. at 
the other end: the bar resting on a fulcrum 9" from 
the weight. (Neglect the weight of the crowbar.) 

Let AB (Fig. 33) be the bar 6 ft. long, and F the 
fulcrum at 9" from A. v Consider moments in inch-lbs. 
about F. Let x be the required force. 






330 


MODERN CARPENTRY 


Moment of 10 cwts. about F=9X10X112 ineh-lbs. 
Moment of x about ^=BFXx=(72—9) X^=63Xx 
inch-lbs. 


, x== 


9X10X112 


63 


=160 lbs. 


Example 2. A man weighing 140 lbs. is using a 
crowbar 5 ft. long. What must be the position of the 
fulcrum to enable him to balance a weight of 1260 lbs. 
at the other end ? 

Let AB be the length of the bar; F the position of 
the fulcrum; and x the length of the long arm in 
inches; then (60—x) is the length of the short arm in 
inches. 

Taking moments about F, 

140x=1260(60—x); 

140x=75600—1260x; 

140x+1260x=75600; 

1400x=75600; 

; * x — jjqq~— 54 in.=4' 6"=length of long arm. 



Fig. 34. 


200lbs. 

J 


Example 3. A lever 7 ft. long is used as shown in 
Fig. 34. If a force of 200 lbs. is applied at P in the 
direction of the arrow, what weight, placed at a point 
12" from the fulcrum, can be raised? 

Taking moments in ft.-lbs. about F, 

WX1=200X7; 

W=1400 lbs. 




MECHANICS OF CARPENTRY 


331 


Loaded Beams. The determination of the proportion 
of the total weight carried by each support of a loaded 
beam—in other words, the upward reaction of each 
support which is necessary to maintain equilibrium— 
affords a good practical example of the theory of paral¬ 
lel forces. 

Example 1. A beam rests upon supports placed 8 
feet apart. A weight of 12 lbs. is placed on the beam 
at a distance of 2 ft. from the right-hand support. 
What proportion of the weight is carried by each of the 
supports, the weight of the beam being neglected? 



6 '. 


Fig. 35. 


*2lbs. 


a 

c 



Arithmetically.—In this case (Fig. 35) the down¬ 
ward force (weight) of 12 lbs. must be balanced by up¬ 
ward forces (reactions) at the points of support, re¬ 
spectively equal to the pressure at these points, and to¬ 
gether equal to 12 lbs.; and the moments of the up¬ 
ward forces about the point c must be equal. 

;. Reaction at AXAc=Reaction at BXBe, 
i.e. Reaction at A :Reaction at B: :Bc :Ac, 

-n, x- , * Sum of reactions _ . 

or Reaction at A: ^ ^ ^ : :Bc :(Bc+Ac), 

i.e. Reaction at A :12 lbs.:: Be :AB. 

This may be expressed in general terms as follows: 

Pressure on one end . that\.^ is ^^ C j r °^ hat . Length between 
caused by any load ’ load ** ot h er end ' supports. 





332 


MODERN CARPENTRY 


therefore Pressure on A : 12 lbs. :: cB : AB; 


.*. Pressure on A = 


12XcB 12X2 
AB 8 


= 3 lbs. 


Similarly, Pressure on B — 


12XAc 12X6 


AB 


6 


= 9 lbs. 


Graphically.—Fig. 36 shows the beam and supports 
with the load in position. The polar diagram is drawn 
as follows: Draw a vertical line a b, representing the 


12 Us. 

A m 3 



Fig. 36. 

weight (12 lbs.) to a suitable scale. From any point 0, 
which may be at any convenient distance from a b, 
draw the triangle 0 a b. Draw as in the figure a verti¬ 
cal line directly under the load, and one under each 
point of support, as Z, x, y. Letter the load A B. and 








333 


MECHANICS OF CARPENTRY 

the space between the supports C. These letters can. 
now be used to denote the reaction at each point of sup- 
P or t i. e., the upward force required to maintain 
equilibrium—which is equal and opposite to the pres¬ 
sure exerted on each support by the load. Anywhere 
in x, as from 1, draw, in the space A, a line 1 2, parallel 
to a 0 ; from 2, in the space B, draw 2 3 parallel te 
b 0. Join 1 3, and through the pole 0, draw 0 c paral¬ 
lel to 3 1. Then ac (on the vertical line of loads ab) 
represents to scale the pressure on the left-hand sup¬ 
port, and c b to the same scale represents the pressure 
on the right-hand support. 


5cwts 


7Cwts. 

O 


V-r*"zr-- 7 ' 


Fig. 37. 


ZCtYts. 



As the reaction at each end is equal in magnitude 
and opposite in direction to the pressure, a c gives the 
amount of the reaction A C, and b c gives the amount 
of the reaction B C; and the sum of the reactions—both 
acting upwards—is equal to the total weight (12 lbs.). 

Example 2. A beam is loaded as shown in Fig. 37. 
Determine the reaction at each end, that is, the upward 
force required at each point of support to maintain 
equilibrium. 


Arithmetically: 

Reaction at A due . , ~ 

to weight ate :wt - atC 


CB : AB; 


. Reaction at A due_wt. at CXCB 5X13 
to weight at C ~~ AB 16 ' 






334 MODERN CARPENTRY 

Similarly, 

Reaction at A due _ wt. at DXDB _ 7X9 
to weight at D AB 16 

Also Reaction at A due _wt. at EXEB_2X2 
to weight at E AB 16 

The total reaction at A is equal to the sum of the 
partial reactions as shown above; or it may be obtained 
directly thus: 

Total reaction at A 

_ (wt. at CXCB) + (wt. at DXDB)+(wt. at EXEB) 

AB 

_(5X13)+(7X9)+(2X2) 132 _ 

16 ~le =8A cwts - 

Similarly, 

Total reaction at B 

_ (wt. at CXCA) + (wt. at DXDA)+(wt. at EXEA) 

AB 

_(5X3)+(7X7)+(2X14) 92 

16 ~16 =5 ' A cwts * 

Graphically.—Construct the vertical line of loads, 
representing to scale the sum of the weights as shown 
in Fig. 38. Fix the pole 0, and draw the dotted lines 
0 a, 0 b, 0 c, 0 d. Letter the loads and draw a dotted 
vertical line directly under each load and under each 
support as shown. From any point in the first line, 
draw in the space A, a line 1 2 parallel to a 0; from 
2 draw 2 3 parallel to b 0; from 3 draw 3 4 parallel to 
c°; and in the space D, from 4 draw 45 parallel to 
d 0. Join 1 to 5, thus completing the funicular poly¬ 
gon. By drawing a line parallel to 5 1— the closing 
line of this polygon—through pole 0, and meeting the 
vertical line of loads at e, it is found that e a equals 








335 


MECHANICS OF CARPENTRY 

the reaction E A, and e d equals the reaction E D; they 
are together equal to the sum of the weights in the 
beam. 



50*is. 4Cwts. 3Cwts. 

,_Q- Q Q o 

-3'. + -3'- i - 3'- * -3'+- 3- 


"K £ 

Fig. 39. 


Example 3. A beam s weighing 6 cwts. is loaded as 
shown in Fig. 39. Determine the reaction at each end 
necessary to produce equilibrium. 












336 


MODERN CARPENTRY 


When the weight of a uniform beam is to be consid¬ 
ered, it may be taken as acting half-way between the 
supports, and thus adding half its weight to each sup- 



port. With this difference the method used is as in the 
previous example. Fig. 40 shows the graphical solu- 
tion. 

















MECHANICS OP CARPENTRY 


337 


Stress Diagrams for Roof Trusses.—Figs. 41 to 50 
show an application of the foregoing graphic methods 
to the determination of stresses in roof trusses. Fig. 
41 is a line diagram of a king-post truss loaded in the 
usual way. It must be noticed that the lettering is ar¬ 
ranged so that every member is indicated by a letter 
on each side. It is first necessary to determine the 
amount of weight carried by each point of support. 
This example is simplified by the symmetrical loading, 
as one half the weight is carried by each point of sup¬ 
port. When this is not the case, the proportion of the 
weight carried by each support must be determined 
first, by a consideration of parallel forces, as in earlier 
examples. 

It is usual when determining the stresses of such a 
truss, to draw the stress diagram, shown in Fig. 46. 
This diagram is a combination of Figs. 42 to 45, which 
are only drawn as separate figures to assist in under¬ 
standing the question more clearly. 

Fig. 42 is the polygon of forces for the joint (1) at 
the foot of the principal rafter on the left. Four forces 
act at the point: A B downwards, B N the principal 
rafter, N G the tie beam, and the upward force. A G— 
the reaction at the point of support. Of these four 
forces the amounts of two, A B and A G, are known; 
it is required to determine the nature and amounts of 
the stress of B N and of N G when acting at the angles 
given. 

Commencing Fig. 42 by drawing a b equal to A B, 
and a g equal to A G, the upward force. As these two 
forces are in the same straight line, and in opposite 
directions, their resultant is the line b g. From b 




338 MODERN CARPENTRY 



Fig. 44. 


Fig. 45. 


















MECHANICS OP CARPENTRY 339 

draw b n parallel to BN; and from g draw g n parallel 
to G N until b n and g n meet. Then b n g is the poly¬ 
gon (a friangle in this case) of forces acting at the 
point; and b n and n g represent the amount of stress 
in the principal rafter and tie-beam respectively. 

The direction of the stress is found by taking the 
forces in order: thus, g b acts upwards; b n acts to¬ 
wards the joint, therefore this member is in compres¬ 
sion ; n g acts from the joint, which indicates that the 
tie-beam is in tension. 

At joint (2), four forces act, namely, B C, B N, C M, 
and N M. The two known forces are B C acting down¬ 
wards and B N towards the joint. For the magnitude 
of the stress on B N has already been found, and its 
direction of action at joint (2) is the opposite to its 
direction at joint (1). Pig. 43 shows the application 
of the polygon of forces to this joint. In it, b c and 
b n are drawn equal and parallel to B C and B N re¬ 
spectively; and, by drawing nm parallel to NM and 
c m parallel to C M, the stress diagram is obtained. 
This shows that the stress in C M, the upper part of the 
principal rafter, is much less than in BN, the lower 
part. By tracing the polygon, it is found that b c acts 
towards the joint, c m towards the joint (therefore C M 
is in compression), m n towards the joint (compression) 
and n b towards the joint (compression as in the pre¬ 
vious figure). 

At joint (3) there are four forces, i. e. CM, CD, 
D L, M L, acting as shown. Of these four forces the 
two C M and C D are known. Since the amount and 
nature of the stress in any member must be the same 
at any intermediate point between the joints, the stress 


340 


MODERN CARPENTRY 





Pig. 49. 


Fig. 50. 





















341 


MECHANICS OP CARPENTRY 

in C M acting upon joint (3) must be as determined by 
the diagram for joint (2). Pig. 44 is the stress dia¬ 
gram; cd is drawn parallel and equal to CD- cm 
parallel and equal to C M; m 1 and d 1 are drawn paral¬ 
lel to M L and L D respectively until they meet. Tak¬ 
ing these forces in order, c d is towards the joint, D L 
towards the joint (compression), LM is from the joint 
(tension), and M C towards the joint (compression). 

The tension stress in LM is caused by the struts 
M N and L H, which transfer part of the loads B C and 
D E respectively to the foot of the king-post. If no 
struts existed in this truss there would'be no stress in 
ML. 

Joint (4) has five forces acting, each one of which 
has already been determined, since the stress diagrams 
for one side of the truss are in this example applicable 
to each side. For example, the diagrams showing the 
stresses in the joints (1) and (2) are applicable to 
(6) and (5) respectively. An examination of Pig. 45 
will show that g n is parallel and equal to GN; n & m is 
parallel and equal to N M; 1 m is parallel and equal to 
LM; and lh being drawn parallel to L H meets mn 
in n, whilst h g is equal to n g. The diagram therefore 
shows the stress in each of the five members. 

In Fig. 46, which is the complete stress diagram for 
the members of the truss, the lettering is identical with 
that in each of the separate Figs. 42 to 45, and will be 
easily understood from them. 

Fig. 47 is a line diagram of a queen-post roof truss, 
and Fig. 48 is the stress diagram of this truss. Simi¬ 
larly , Figs. 49 and 50 are respectively the line diagram 
of and the stress diagram for, a composite roof truss, 


342 


MODERN CARPENTRY 


sometimes named a German truss. The detailed ex¬ 
planation already given will enable the figures to be 
understood. 


STRENGTH OF WOODEN BEAMS. 

For the purpose of calculating the carrying capacity 
of wooden beams, it is necessary to notice the nature of 
the stresses to which they are subjected, as well as the 
manner in which they are loaded, and the arrangement 
of the load. 

Stress and Strain. When a weight, or other force, 
acts upon a beam, it tends to change the shape and 
size of the beam. The force is technically called a 
stress, while the change in shape or size is called a 
strain. When a beam or girder, supported at both 



Fig. 51. 


ends, is loaded, the upper part tends to shorten. The 
lower fibres, on the other hand, are in a state of ten¬ 
sion, as they tend to stretch. The force acting on the 
upper fibres of such a beam is therefore a compression 
stress; that on the lower fibres is a tension stress. 

The existence of these stresses may be made very 
apparent either by making a saw-cut across, or by 
actually cutting out a wedge-shaped piece from, the 
middle of a beam of wood for half its depth, as shown 
in Fig. 51. On resting the beam on two supports with 
the cut edge uppermost, and then loading it, it will be 






MECHANICS OF CARPENTRY 


343 


seen that the saw-cut closes. This shows that the fibres 
on the upper side are in a state of compression. If the 
same beam is now turned over so that the saw-cut is on 
the lower side, and again loaded, the tendency is for 
the cut to open, thus showing that the fibres on the 
lower side are in a state of tension. 



Fig. 52. 


Shearing Stresses. A shearing stress is one which 
gives the fibres of the wood a tendency to slide over 
one another. A shearing stress may be either in the 
direction of the fibres or at right angles to them. To 
illustrate a shearing stress in the direction of the fibres 
of a beam, imagine the beam cut into a number of 



boards; place these on the top of each other in the 
position of a beam resting upon supports at each end, 
and place a load in the middle. The result will be that 
the beam will bend as shown in Fig. 52, and the boards 
will slide over each other. A shearing stress across 
the fibres of a loaded beam can be illustrated by tak¬ 
ing a bar of soap, or some such soft material, resting 
it upon supports, and loading it. The result will be as 
shown in Fig. 53. 







344 


MODERN CARPENTRY 


Methods of Arranging Beams. The nature and 

amount of stress in the fibres of a loaded beam depend 
upon the way the beam is supported and on the ar¬ 
rangement of the load. Thus a cantilever is a beam 
with one end only secured upon a support, the other 
end overhanging. The load upon a cantilever may be 
a concentrated load at the outer end, as in Fig. 57, or 
the load may be anywhere between the outer end and 

Tension 

-- ..q 

Compression; — - j 

Fig. 54. 

the supported end; a number of loads of varying 
weights may be distributed over the length; the load 
may be a uniformly distributed one extending over the 
length of the beam, or it may be a combination of a 
concentrated load and a distributed load. A cantilever 
loaded in any of the ways just described has the fibres 
in the upper edge in a state of tension, those in the 
lower half being in compression (Fig. 54). 

Compression/ 

~ -'tension/ - 

Fig. 55. 

A beam supported at both ends may be loaded in any 
of the ways described for the cantilever, with the re¬ 
sult that the stresses will be as shown in Fig. 55 ; i. e . 
the upper part will be in compression, and the lower 












MECHANICS OF CARPENTRY 345 

half in a state of tension. The stresses in the various 
parts of a loaded beam which has the ends fixed differ 
from those of the beam which simply rests upon sup¬ 
ports. They are illustrated in Fig. 56, which shows 
that for a distance of about one-fourth from each end 
the beam takes the form of a cantilever, and has the 
fibres in the upper half in a state of tension and the 
lower fibres in compression. The remainder of the 
beam has the upper fibres in compression and the 
lower part in tension. The neutral axis of all these 
beams is in the centre of the depth. If a long beam has 
intermediats supports as in Fig. 56, it may be regarded 
as being “fixed” at the points of intermediate support. 


~p-r-T"' 




r ' 


Fig. 56. 


Bending Moments. For the purpose of making com¬ 
parisons of the relative strengths of loaded beams, a 
further consideration of the “moment of a force” is 
necessary. Since the tendency to bending, to which a 
given beam is subject at any point, depends upon the 
moments of the stresses about that point, it is obvious 
that the relative strengths of beams may be measured 
in terms of moments. The bending moment at any 
given section is the algebraic sum of all the external 
forces acting on one side of the section. Since it is at 
the point where the greatest bending moment occurs 
that the beam is subjected to the greatest stress, it fol¬ 
lows that it is of some importance to be able to deter- 






346 


MODERN CARPENTRY 


mine the bending moment of beams loaded under dif¬ 
ferent conditions. The bending moment—like other 
moments—must always be expressed in terms of a 
length and a force. 

Example 1. A cantilever carries a load of 6 tons at 
its outer end, which is 5 ft. from the supporting wall. 
Determine the maximum bending moment, and also the 
bending moment at 2 ft. from the wall. 



The greatest tendency to bending will be at the point 
of support, i. e. at a distance of 5 ft. from the load. 

Maximum bending moment=5X6=30 ft.-tons. 

Bending moment at 2 ft. from the wall (i. e. 3 ft. 
from the load) =3X6=18 ft.-tons. 

The bending moment at any distance from the load 
may be determined graphically by drawing, as in Fig. 
57, a vertical line AB 30 units long (respecting the 
maximum bending moment) under the point of support 
A (i. e. the point where the bending moment is a maxi¬ 
mum), and joining B C. Then the bending moment at 
the point a will be represented by the length’ of a b 
drawn parallel to A B. 







MECHANICS OF CARPENTRY 347 

Example 2. A cantilever projects 4 ft. and carries a 
uniformly distributed load of 8 cwts. along the upper 
edge. Determine the maximum bending moment, and 
also draw a diagram from which the bending moment 
at any section along the length of the cantilever may 
be determined. 



Fig. 58. 


A load arranged as'shown in Fig. 58 is equivalent 
to a concentrated load of 8 cwts. acting in the middle 
of the length, i. e. 2 ft. from the point of support. The 
maximum bending moment will therefore be 

8X2=16 ft.-cwts. 

Fig. 58 is the diagram from which the bending mo¬ 
ment at any section may be determined. The load is 
supposed divided into 4 equal parts, and the bending 
moment due to each part is drawn to scale on the verti- 









348 


MODERN CARPENTRY 


cal line A E. The weight of Z acts at 3' 6" from A, 
and the maximum bending moment due to Z=2X3.5= 
7 ft.-cwts. Draw AB 7 units long. Similarly, the 
maximum bending moment due to Y=2X2.5=5 ft.- 
cwts., and is represented by BC; maximum bending 


Fig. 59. 


Fig. 60. 





moment due to X=2X 1.5=3 ft.-cwts., represented by 
CD; maximum bending moment due to W=2X0.5=1 
ft.-cwt., represented by D E. The maximum bending 
moment due to the total load is therefore 7+5+3+1 
~16 ft.-cwts., and is represented by A E. Draw a ver¬ 
tical line through the centre of each part of the load, 
and complete the triangles A a B, B b C, C c D, D d E. 
Draw an even curve touching the lines E d, D c, C b, 
B a. This curve is a parabola. The bending moment 






















MECHANICS OF CARPENTRY 349 

at any section P is represented by the length of the 
vertical line P Q cutting the parabola at Q. 

The following formula- are used for determining the 
relative bending moments (and therefore the relative 
strengths required) of beams loaded in various ways. 
In each case L=the length of the beam and W—the 
weight of the load. 


Cantilever fixed at one end and loaded 
at the other end (Fig. 57), 

Cantilever fixed at one end and loaded 
with uniformly distributed load, 

Beam supported at both ends and 
loaded with a central load (Fig. 59), 

Beam supported at both ends and 
loaded with a uniformly distrib¬ 
uted load (Fig. 65), 

Beam fixed at both ends and loaded 
with a central load (Fig. 62), 

Beam fixed at both ends and loaded 
with a uniformly distributed load, 


Maximum 

Bending 

Moment. 

J WL 

l WL 
1 2 

I 5 ? 

1 WL 

J 8 

) WL 

i 8 

l WL 
J 12 


Relative 

Strength. 

1 

2 

4 

8 

8 

12 


Figs. 60 to 63 show beams loaded in various ways, 
and serve to illustrate the method of determining 
graphically the bending moment at any section of the 
beam.- 


It will be noticed that the maximum bending moment 
of a beam supported at each end is in each case in the 
line of the load, and with a central or an evenly dis¬ 
tributed load is at the middle of the length of the beam. 

Calculation of the Transverse Strength of Wooden 
Beams. Other things being equal, the strength of a 


350 


MODERN CARPENTRY 


rectangular wooden beam is directly proportional to 
the breadth in inches multiplied by the square of the 
depth in inches, and inversely proportional to the 
length in feet. Of course the nature of the material 
is also an important factor, since timber, even of the 
same kind, varies in strength to a considerable extent. 
Each beam therefore has what is called a natural con¬ 
stant, which must be considered in the calculation of 
its carrying capacity. To obtain this constant, it is 
usual to take a bar of similar wood, 1 inch square in 
section, and long enough to allow of its being placed 
on supports 1 foot apart. The constant is the weight 
of the central load, which is just sufficient to break 
the bar. The constant may be expressed in lbs., cwts., 
tons, etc., and the carrying capacity will always be in 
the same terms. The following constants (in cwts.) 
may be adopted for the purposes of calculation: oak, 
ash and pitch pine, 5; red deal, red pine and beech, 4; 
white and yellow pine, 3. 

Another important consideration is the ratio which 
the breaking load of a beam bears to the “safe” load. 
This ratio is called the factor of safety, and its value 
depends upon whether the load is a live—a constantly 
moving—load, or a dead (i. e., a stationary) load. The 
factor of safety for a dead load is usually taken at 5, 
which means that the safe load upon a beam must not 
exceed one-fifth of the breaking load; the factor of 
safety for a live load is often taken at 10. 

For beams supported at both ends the formula 

W = ^iT may be used for the Proses of calcula¬ 
tion when; 


351 


MECHANICS OF CARPENTRY 


W—breaking weight or maximum carrying capacity 
of a centrally loaded beam, expressed in the 
same terms as the constant, 
b=breadth of the beam in inches, 
d=depth of the beam in inches, 

L=length of the beam in feet, 
c the constant, found by experiment as described 
above, and expressed in terms of lbs. or cwts. 
To illustrate the above formula, take two pieces of 
the same kind of wood say 7 ft. long, 6 in. wide and 2 
in. thick. Place one of these pieces flat, and the other 
on one edge, the distance between the supports in each 
case being 6 ft. As the constant is the same in both 
(say 5 cwts.), the carrying capacity of each will be 


expressed by the formula W : 


bd 2 c . 

= ~IT’ 


for the flat beam, w= — X2X5 =20, 


for the one on edge, W= —^ X ^ X ^ 60; 


and the relative strengths will be as 20:60 or as 1:3. 
When it is necessary to find other terms than W, 

the equation W=^-?may be expressed as follows: 



WL 
d 2 c 5 


d 2 = 


WL 
be ’ 




WL 

bd 2 


The value of W for a distributed load is twice that 

for a concentrated load, i. e., W=— T d e . When the 
, L 

ends are fixed the carrying capacity is increased by 
about one-half. 





352 


MODERN CARPENTRY 


For safe central loads the formula W = is used: 

LF 

F being the factor of safety. 

Example 1. Find the maximum carrying capacity 
of a centrally loaded wooden beam of pitch pine, 13 
ft. long (12 ft. between the supports), 10 in. wide, 
and 6 in. thick, (1) when placed on edge; (2) when 
placed flat. Assume a constant of 5 cwts. 

Applying the formula 

T T T_ b d 2 c, ^exioxioxs _ 

W— k (1) W- — -=250 cwts. when on edge. 


(2) W= 


12 

10X6X6X5 


—150 cwts. when placed flat. 


Example 2. What would be the maximum safe load 
to which the beam in Ex. 1 may be subjected (1) as a 
central load; (2) as a uniformly distributed load? 

Formula for safe central load using a factor of safety 
of 5, is 


Safe central load= 


bd 2 c 

LF 


_ 6X10X10 X5 

12X5 

10X6X6X5 


► 

=50 cwts. for beam on edge, 


12X5 cwfcs * for beam placed flat. 

Formula for safe central load using a factor of safety 
using factor of safety of 5, is 

2bd 2 c 


Safe distributed load c 


LF 


_ 2X 6X10X10X5 

12X5 100 cwts. for beam on edge, 

_2X10X6X6X5 

12X5 —60 cwts. for beam placed flat 


or 









353 


mechanics op carpentry 

Example 3. Find the breadth of a beam of oak rest¬ 
ing upon supports 18 feet apart, the beam being 12 in. 
deep, to carry safely a uniformly distributed load of 
5 tons. Constant 5 cwts. 


Safe distributed load, W c . 


• h —WLF_ (5X20)X18X5 25 

2d 2 c 2X12X12X5 ~J~ 6A lnches - 

Example 4. A beam of red or yellow pine 20 ft. 
long (between supports), and 10 in. broad has to carry 
safely (1) as a central load, (2) a distributed load of 4 
tons. What must be the minimum depth of the beam 
in each case? (Constant 4 cwts.) 


WLF 
be > 


With a central load d 2 =* 




(80X20X5 - 

10X4 Y200—14.14 inches. 


. d ; 


With a distributed load 

/WLF_ /80X20X5 — 

d ~ \ 2be 2X10X4 ~ V:100=10 in ches. 

Example 5. What size of beam is required to carry 
safely a central load of 35 cwts. oyer a 10 ft. span; 
the depth and breadth of the beam being in the pro¬ 
portion of 7:5 ? (Constant 5 cwts.) 


c 



WLF 












354 


MODERN CARPENTRY 


5d 3 _WLF 
7 c 


„ 7WLF 7X35X10X5 
d 3 =—t =400. 


5c 


5X5 


d=V 490=7.88 //= nearly 8". 


. b= 5X7 1 88 


=5.6 inches. 


The strength of hitched girders may be calculated 
by considering the wooden beam and iron flitch sepa¬ 
rately. The thickness of the flitch is usually about 
1-12 that of the wooden beam. The constant for 
wrought iron is 25 cwts. 

Deflection. In arranging beams it is necessary to 
consider not merely the strength of the beam, but also 
its liability or otherwise to be bent out of shape—or 
deflected by the load placed upon it; since a beam 
which is overloaded and bent to a large extent has the 
fibres strained and therefore permanently weakened. 
The resistance which a beam offers to deflection is 
called its stiffness. It should be noticed that the 
“strongest” beam is not necessarily the “stiffest,” nor 
the stiffest beam the strongest. 

It is of importance to be able to determine the cross- 
sections of the strongest and the stiffest beams respec¬ 
tively which can be cut from a given log. 

Suppose the log is of circular cross-section. 

(a) To find the cross-section of the strongest beam 
—Draw a diameter A B(Fig. 64) and divide it into 3 
equal parts at 1 and 2. From 1 and 2 draw perpen¬ 
diculars to AB cutting the circumference at C and D 
respectively. Join ACBD. The rectangle ACBD is the 
section required. 





.355 


MECHANICS OF CARPENTRY 

(b) To find the cross-section of the stiffest beam — 
Divide the diameter AB (Fig. 65) into 4 equal parts 
at 1, 2, 3, and draw 1C and 3D perpendicular to AB, 
and cutting the circumference in C and D respectively! 
The rectangle ACBD is the section required. 

Since the strength of a beam is proportioned to 
bd 2 

L , and the value of this fraction increases as d 
increases when bd (i. e'., the sectional area) remains 



c* 


/ 

D 


Fig. 64. 


Fig. 65. 


constant, the strongest beam of any given sectional 
area would be that of greatest depth if the tendency 
to buckling could be avoided. In the case of floor 
joists the ratio of depth to breadth is often as 3 :1 or 
even 4:1, and the tendency to buckling is overcome by 
strutting. The strongest beam is that which has the 
depth to the breadth as 7:5. 







356 


MODERN CARPENTRY 


Pulleys. It is necessary to consider one or two sim¬ 
ple arrangements by which pulleys are used for hoist¬ 
ing purposes. In the following examples the friction 
will for the sake of simplicity be neglected, although 
in practice it must be taken into account. Fig. 66 
illustrates the simplest application of the pulley. It is 



Fig. 66. 



plain that when the forces acting on the pulley are in 
equilibrium they are equal, and the only advantage 
gained is in the change of direction of the force re¬ 
quired to balance W. Therefore in this example P=W. 

In Fig. 67 the force balancing P is the tension of 
the cord A, which is equal to that of B. The sum of 













MECHANICS OF CARPENTRY .357 

these two equal tensions is plainly equal to the weight 

W. Therefore P= and the mechanical advantage 
is 2 . 

Figs. 68 and 69 are illustrations of a two- and three- 
sheaved pulley block respectively. By arranging pul¬ 
leys side by side in this manner, and using a combina¬ 
tion of two similar blocks as in Fig. 70 a mechanical 
advantage equal to the number of pulleys around 
which the rope passes is obtained. In other words, the 


Fi g. 68. Fig. 69> 


power required is equal to the weight raised divided 
by the number of pulleys around which the rope passes. 
Thus with 3 pulleys in each block there will be six 
cords, and the power required to balance a weight of 
18 cwts. will be 18 -f- 6 =' 3 cwts., plus the force re¬ 
quired to overcome friction. 





358 


MODERN CARPENTRY 


W 


Fig. 70. 


Specific Gravity. The specific 
gravity, or relative density, of a 
body is the ratio of the weight of 
that body to the weight of an equal 
volume of water. Thus a block of 
wood weighing 40 lbs. per cubic foot 
40 

has a specific gravity of-^^r?-— 0.64 
oZ.o 

(since a cubic foot of water weighs 
62.5 lbs.). 

When a body floats in water, and 
is therefore in equilibrium, the 
weight of the body is balanced by 
an equal upward reaction, the 
weight of the water displaced being 
equal to the total weight of the 
floating body. 

Example. A block of wood 9"x9" 
x9 , floats in water with its upper 
surface 2.5" above the surface of the 
water. Find its specific gravity. 

6 13 

= jg of the block is submerged. 

By definition, the specific gravity 
of the wood is the ratio of the 
weight of any portion of the block 


to the weight of an equal volume of water. Consider 
the part of the block below the surface of the water. 

Specific Weight of Emerged part of wood 

Weight of displaced water 

N _ Weight of submerged part of wood 

Weight of whole block 















MECHANICS OF CARPENTRY 


35 $ 


_ Volume of submerged part of woo d 

Volume of whole block 

= 1 = 0 ,, 

In addition to the foregoing I have thought it es¬ 
sential to add the following special notes which are 
the results of actual experiments, and which will, I 
am sure, appeal to the American workmen, especially 
those who are located in the West, where timber is yet 
often employed in heavy structures. 

STRENGTH OF TIMBER. 

Proposition I. The strengths of the different pieces 
of timber, each of the same length and thickness, are 
in proportion to the square of the depth; but if the 
thickness and depth are both to be considered, then 
the strength will be in proportion to the square of the 
depth, multiplied into the thickness; and if all the 
three dimensions are taken jointly, then the weights 
that will break each will be in proportion to the square 
of the depth multiplied into the thickness, and divided 
by the length; this is proved by the doctrine of me¬ 
chanics. Hence a true rule will appear for proportion¬ 
ing the strength of timbers to one another. 

Rule. Multiply the square of the depth of each piece 
of timber into the thickness; and each product being 
divided by the respective lengths, will give the pro¬ 
portional strength of each. 

Example. Suppose three pieces of timber, of the 
following dimensions: 

The first, 6 inches deep, 3 inches thick, and 12 feet 

long. 



360 


MODERN CARPENTRY 


The second, 5 inches deep, 4 inches thick, and 8 feet 
long. 

The third, 9 inches deep, 8 inches thick, and 15 feet 
long. The comparative weight that will break each 
piece is required. 

Ans. 9, 12%, and 43 1-5. 

Therefore the weights that will break each are 
nearly in proportion to the numbers 9, 12, and 43, 
leaving out the fractions, in which you will observe, 
that the number 43 is almost 5 times the number 9; 
therefore the third piece of timber will almost bear 5 
times as much weight as the first, and the second piece 
nearly once and a third the weight of the first piece; 
because the number 12 is once and a third greater 
than the number 9. 

The timber is supposed to be everywhere of the same 
texture, otherwise these calculations cannot hold true. 

Proposition II. Give the length, breadth, and depth 
of a piece of timber; to find the depth of another piece 
whose length and breadth are given, so that it shall 
bear the same weight as the first piece, or any number 
.of times more. 

Rule. Multiply the square of the depth of the first 
piece into its breadth, and divide that product by its 
length: multiply the quotient by the number of times 
as you would have the other piece to carry more weight 
than the first, and multiply that by the length of the 
last piece, and divide it by its width; out of this last 
quotient extract the square root, which is the depth 
required. 

Example I. Suppose a piece of timber 12 feet long, 

6 inches deep, 4 inches thick; another piece 20 feet 


361 


MECHANICS OF CARPENTRY 

long, 5 inches thick; required its depth, so that it shall 
bear twice the weight of the first piece. 

Ans. 9.8 inches, nearly. 

Example II. Suppose a piece of timber 14 feet long, 
8 inches deep, 3 inches thick; required the depth of 
another piece 18 feet long, 4 inches thick, so that the 
last piece shall bear five times as much weight as the 
first. 

Ans. 17.5 inches, nearly. 

Note.—As the length of both pieces of timber is 
divisible by the number 2, therefore half the length 
of each is used instead of the whole; the answer will 
be the same. 

Proposition III. Given the length, breadth, and 
depth, of a piece of timber; to find the breadth of 
another piece whose length and depth is given, so that 
the last piece shall bear the same weight as the first 
piece, or any number of times more. 

Rule. Multiply the square of the depth of the first 
piece into its thickness; that divided by its length, 
multiply the quotient by the number of times as you 
would have the last piece bear more than the first; 
that being multiplied by the length of the last piece, 
and divided by the square of its depth, this quotient 
will be the breadth required. 

Example I. Given a piece of timber 12 feet long, 

6 inches deep, 4 inches thick; and another piece 16 feet 
long, 8 inches deep; required the thickness, sc that it 
shall bear twice as much weight as the first piece. 

Ans. 6 inches. 

Example II. Given a piece of timber 12 feet long, 

5 inches deep, 3 inches thick; and another piece 14 


362 


MODERN CARPENTRY 


feet long, 6 inches deep; required the thickness, so 
that the last piece may bear four times as much weight 
as the first piece. 

Ans. 9.722 inches. 

Proposition IV. If the stress does not lie in the 
middle of the timber, but nearer to one end than the 
other, the strength in the middle will be to the strength 
in any other part of the timber, as 1 divided by the 
square of half the length is to 1 divided by the rect¬ 
angle of the two segments, which are parted by the 
weight. 

Example I. Suppose a piece of timber 20 feet long, 
the depth and width is immaterial; suppose the stress 
or weight to lie five feet distant from one of its ends, 
consequently from the other end 15 feet, then the above 


portion will be —-— M Hf . 

10 X 10 100 5 X 15 

strength at five feet from the end is to the strength 


100 


= ^ as the 
75 


at the middle, or ten feet, or as 


100 

100 


= 1 


100 

75 


3* 


Hence it appears that a piece of timber 20 feet long 
is one-third stronger at 5 feet distance for the bearing, 
than it is in the middle, which is 10 feet, when cut in 
the above proportion. 

Example II. Suppose a piece of timber 30 feet long; 
let the weight be applied 4 feet distant from one end, 
or more properly from the place where it takes its 
bearing, then from the other end it will be 26 feet, 


and the middle is 15 feet; 


then, 


1 1 _ 1 _ 

15 X 15 225 : 4x26 ~ 




1 


225 


225 


17 






MECHANICS OP CARPENTRY 


363 


Hence it appears that a piece of timber 30 feet long 
will bear double the weight, and one-sixth more, at 
four feet distance from one end, than it will do in the 
middle, which is 15 feet distant. 

Example III. Allowing that 266 pounds will break 
a beam 26 inches long, requireth the weight that will 
break the same beam when it lies at 5 inches from 
either end; then the distance to the other end is 21 
inches; 21X5=105, the half of 26 inches is 3:13X13 
=169; therefore the strength at the middle of the 
piece is to the strength at 5 inches from the end, as 
169 169 v 169 , 

169 : 105 or as 1 : 105 the P r °Portion is stated thus : 1 : 
169 

: : 266 : 428-f, Ans. 

From this calculation it appears, that rather more 
than 428 pounds will break the beam at 5 inches dis¬ 
tance from one of its ends, if 266 pounds will break 
the same beam in the middle. 

By similar propositions the scantlings of any timber 
may be computed, so that they shall sustain any given 
weight; for if the weight one piece will sustain be 
known, with its dimensions, the weight that another 
piece will sustain, of any given dimensions, may also 
be computed. The reader must observe, that although 
the foregoing rules are mathematically true, yet it is 
impossible to account for knots, cross-grained wood, 
&c., such pieces being not so strong as those which are 
straight in the grain; and if care is not taken in 
choosing the timber for a, building, so that the grain 
of the timbers run nearly equal to one another, all 
rules which can be laid down will be baffled, and con- 


364 


MODERN CARPENTRY 


sequently all rules for just proportion will be useless 
in respect to its strength. It will be impossible, how¬ 
ever, to estimate the strength of timber fit for any 
building, or to have any true knowledge of its propor¬ 
tions, without some rule; as without a rule everything 
must be done by mere conjecture. 

Timber is much weakened by its own weight, except 
it stands perpendicular, which will be shown in the 
following problems; if a mortice is to be cut in the 
side of a piece of timber, it will be much less weak¬ 
ened when cut near the top, than it will be if cut at 
the bottom, provided the tenon is driven hard in to 
fill up the mortice. 

The bending of timber will be nearly in proportion 
to the weight that is laid on it; no beam ought to be 
trusted for any long time with above one-third or one- 
fourtli part of the weight it will absolutely carry 
tor experiment proves, that a far less weight will 
break a piece of timber when hung to it for any con¬ 
siderable time, than what is sufficient to break it when 
first applied. 


Problem I. Having the length and weight of a beam 
that can just support a given weight, to find the length 
or another beam of the same scantling that shall just 
break with its own weight. 


Let l=the length of the first beam, 

L=the length of the second; 
a =the weight of the first beam, 
w=the additional weight that will break it. 

And because the weights that will break beams of the 
same scantling are reciprocally as their lengths, 


MECHANICS OF CARPENTRY 


365 


, 1.1 a w '2 

therefore j- . ^ : : w+- : - ^ l = W = the weight that 

will break the greater beam; because w +| is the whole 
weight that will break the lesser beam. 

But the weights of beams of the same scantling are 
to one another as their lengths : 

Whence, 1 : L : : —=W half the weight of the 

greater beam. 


Now the beam cannot break by its own weight, 
unless the weight of the beam be equal to the weight 
that will break it: 

w + - 

Wherefore, ~ = —jr-^1 = L 2 a = 2wl 2 +al 2 , 


a: 2w-f a : : l 2 : L 2 , consequently VL 2 = L = the length 
of the beam that can just sustain its own weight. 

Problem II. Having the weight of a beam that can 
just support a given weight in the middle, to find the 
depth of another beam similar to the former, so that 
it shall just support its own weight. 

Let d=tlie depth of the first beam ; 
x=the depth of the second; 
a=the weight of the first beam; 
w=the additional weight that will break the first 
beam; 


then will w+tor= the whole weight that will 
Z A 


break the lesser beam. 





366 


MODERN CARPENTRY 


And because the weights that will break similar beams 
are as the squares of their lengths, 

• . 2 . . 2w +a # 2 x 2 Xw+ax 2 

.2 * 2d 2 “ W 

the weights of similar beams are as the cubes of their 
corresponding sides: 

Hence d 3 : x 3 : : § : 

2 2d 3 

. ax 3 2x 2 w-fx 2 a . _ „ 

• • 2^= 2d 2 -* ' ax = 2wd + ad 

. . a : a+2w : : d : x = the depth required. 

As the weight of the lesser beam is to the weight of 
the lesser beam together with the additional weight, 
so is the depth of the lesser beam to the depth of the 
greater beam. 

Note—Any other corresponding sides will answer 
the same purpose, for they are all proportioned to one 
another. 

Example. Suppose a beam whose weight is one 
pound, and its length 10 feet, to carry a weight of 
399.5 pounds, required the length of a beam similar 
to the former, of the same matter, so that it shall break 
with its own weight, 
here a=l 
and w=399.5 

then a+2 w=800=l+2X399.5 
d=10 

Then by the last problem it will be 1:800:10: 8000 
~ x for the length of a beam that will break by its own 
weight. 





MECHANICS OF CARPENTRY 367 

Problem HI. The weight and length of a piece of 
timber being given, and additional weight that will 
break it, to find the length of a piece of timber similar 
to the former, so that this last piece of timber shall be 
the strongest possible: 

Put l=the length of the piece given 
w=half its weight, 

W=the weight that will break it; 
x=the length required. 

Then, because the weights that will break similar pieces 
of timber are in proportion to the squares of their 
lengths, 

• 12 . —2 TTT Wx 2 -fwx 2 

. . 1 .X . : W+w : p -= the whole weight that 

breaks the beam; 

and because the weights of similar beams are as the 
cubes of their lengths, or any other corresponding 
sides, 

then l 3 : x 3 : : w : “pp the weight of the beam; 
Wx 2 +wx 2 wx 3 

consequently p less “p~is the weight that breaks 
the beam= a maximum; therefore its flexion is nothing. 

that is, 2Wxx-f 2wxx—nothing. 

2 W+ 2 w = therefore, x = I X — 0 +2w 

1 3w 

Hence it appears from the foregoing problems, that 
large timber is weakened in a much greater propor¬ 
tion than small timber, even in similar pieces, there- 






368 


MODERN CARPENTRY 


fore a proper allowance must be made for the weight 
of the pieces, as I shall here show by an 

Example. Suppose a beam 12 feet long, and a foot 
square, whose weight is three hundred pounds, to be 
capable of supporting 20 hundred weight, what weight 
will a beam 20 feet long, 15 inches deep, and 12 thick, 
be able to support? 


12 square inches 

15 

12 

15 

144 

75 

12 

15 

12) 1728 

225 

144 

12 

2.0)270.0 

135 


But the weights of both beams are as their solid con¬ 
tents : 


therefore 12 inches square 
12 .. 

144 

144 inches=12 feet long 


576 

576 

144 


20736 solid contents of the 1st beam 











MECHANICS OF CARPENTRY 


369 


15 deep 
12 wide 

180 

240 length in inches 


7200 

360 


43200 solid contents of the 2d beam 


20736 :43200 : :3 
3 

-cwt. lb. 

20736)129600(6.. 28= the weight of the 2d beam 
124416 


5184 

112 


10368 

5184 

5184 


20736)580608(28 

41472 


165888 

165888 










370 


MODERN CARPENTRY 


144: :135 : :21.5 by prop. 1. 
21.5 


67.5 

135 

270 


12)2902.5 

12)241.875 


20.25625 

112 


31250 

15625 

15625 


17.50000 

16 


30 

5 


8.0 


21 cwt. 56 lbs. is the weight that will break the first 
beam, and 20 cwt. 17 lb. 8 oz. the weight that will 
break the second beam; deduct out of these half their 
own weight. 

20: :17: :8 
3 : :14 : :0 half 


17...3..8 


Now 20 cwt. is the additional weight that will break 
the first beam; and 17 cwt. 3 lb. 8 oz. the weight that 











V 

MECHANICS OF CARPENTRY 371 

will break the second: in which the reader will ob¬ 
serve, that 10: :3: :8 has a much less proportion to 20, 
than 20 cwt. 17 lb. 8 oz. has to 21: :46. From these 
examples, the reader may see that a proper allowance 
ought to be made for all horizontal beams; that is, half 
the weights of beams ought to be deducted out of the 
whole weight that they will carry, and that will give 
the weight that each piece will bear. 

If several pieces of timber of the same scantling and 
length are applied one above another, and supported 
by props at each end, they will be no stronger than if 
they were laid side by side; or this, which is the same 
thing, the pieces that are applied one above another 
are no stronger than one single piece whose width is 
the width of the several pieces collected into one, and 
its depth the depth of one of the pieces; it is there¬ 
fore useless to cut a piece of timber lengthways, and 
apply the pieces so cut one above another, for these 
pieces are not so strong as before, even if bolted. 

Example. Suppose a girder 16. inches deep, 12 in¬ 
ches thick, the length is immaterial, and let the depth 
be cut lengthways in two equal pieces; then will each 
piece be 8 inches deep, and 12 inches thick. Now, 
according to the rule of proportioning timber, the 
square of 16 inches, that is, the depth before it was 
cut, is 256, and the square of 8 inches is 64; but twice 
64 is only 128, therefore it appears that the two pieces 
applied one above another, are but half the strength 
of the solid piece, because 256 is double 128. 

If a girder be cut lengthways in a perpendicular 
direction, the ends turned contrary, and then bolted 
together, it will be but very little stronger than before 


372 


MODERN CARPENTRY 


it was cut; for although the ends being turned give to 
the girder an equal strength throughout, yet wherever 
a bolt is, there it will be weaker, and it is very doubt¬ 
ful whether the girder will be any stronger for this 
process of sawing and bolting; and I say this from ex¬ 
perience. 

If there are two pieces of timber of an equal scant¬ 
ling, the one lying horizontal, and the other inclined, 
the horizontal piece being supported at the points 
e and f, and the inclined piece at c and d, perpendi¬ 
cularly over e and f, according to the principles of 
mechanics, these pieces will be equally strong. But, 
to reason a little on this matter, let it be considered, 
that although the inclined piece D is longer, yet the 
weight has less effect upon it when placed in the 
middle, than the weight at h has upon the horizontal 
piece C, the weights being the same; it is therefore 
reasonable to conclude, that in these positions the one 
will bear equal to the other. 

The foregoing rules will be found of excellent use 
when timber is wanted to support a great weight; for, 
by knowing the superincumbent weight, the strength 
may be computed to a great degree of exactness, so 
that it shall be able to support the weight required. 
The consequence is as bad when there is too much tim¬ 
ber, as when there is too little, for nothing is more 
requisite than a just proportion throughout the whole 
building, so that the strength of every part shall al¬ 
ways be in proportion to the stress; for when there is 
more strength given to some pieces than others, it 
encumbers the building, and consequently the founda¬ 
tions are less capable of supporting the superstructure. 


MECHANICS OF CARPENTRY 


373 


No judicious person, who has the care of construct¬ 
ing buildings, should rely on tables of scantling, such 
as are commonly in books; for example, in story posts 
the scantlings, according to several authors, are as fol¬ 
lows: 

For 9 feet high 6 inches square 

12- 8 

15-10 

18-12 

Now, according to this table, the scantlings are in¬ 
creased in proportion to the height; but there is no pro¬ 
priety in this, for each of these will bear weight in 
proportion to the number 9, 16, 25, and 36, that is, in 
proportion to the square of their heights, 36 being 4 
times 9; therefore the piece that is 18 feet long, will 
bear four times as much weight as that piece which 
is 9 feet long; but the 9 feet piece may have a much 
greater weight to carry than an 18 feet piece, suppose 
double: in this case it must be near 12 inches square 
instead of 6. The same is also to be observed in 
breast-summers, and in floors where they are wanted 
to support a great weight; but in common buildings, 
where there are only customary weights to support, 
the common tables for floors will be near enough for 
practice. 

To conclude the subject, it may be proper to notice 
the following observations which several authors have 
judiciously made, viz.: that in all timber there is 
moisture, wherefore all bearing timber ought to have 
a moderate camber, or roundness on the upper side, 
for till that moisture is dried out, the timber will swag 
with its own weight. 





374 


MODERN CARPENTRY 


But then observe, that it is best to truss girders when 
they are fresh sawn out, for by their drying and 
shrinking, the trusses become more and more tight. 

That all beams or ties be cut, or in framing forced 
to a roundness, such as an inch in twenty feet in 
length, and that principal rafters also be cut or forced 
in framing, as before; because all joists, though ever 
so well framed, by the shrinking of the timber and 
weight of the covering will swag, sometimes so much 
as not only to be visible, but to offend the eye: by 
this precaution the truss will always appear well. 

Likewise observe, that all case-bays, either in floors 
or roofs, do not exceed twelve feet if possible; that is, 
do not let your joints in floors exceed twelve feet, nor 
your purlins in roofs, &c., but rather let their bearing 
be eight, nine, or ten feet. This should be regarded 
in forming the plan. 

Also, in bridging floors, do not place your binding 
or strong joists above three, four, or five feet apart, 
and take care that your bridging of common joists are 
not above twelve or fourteen inches apart, that is, be¬ 
tween one joist and another. 

Also, in fitting down tie-beams upon the wall plates, 
never make your cocking too large, nor yet too near 
the outside of the wall plate, for the grain of the wood 
being cut across in the tie-beam, the piece that remains 
upon its end will be apt to split off, but keeping it 
near the inside will tend to secure it. 

Likewise observe, never to make double tenons for 
bearing uses, such as binding joists, common joists, or 
purlins; for, in the first place, it very much weakens 
whatever you frame it into, and in the second place, 


MECHANICS OF CARPENTRY 


375 


it is a rarity to have a draught to both tenons, that is, 
to draw both joists close; for the pin in passing through 
both tenons, if there is a draught in each, will bend 
so much, that unless it be as tough as wire, it must 
needs break in driving, and consequently do more hurt 
than good. 

Roots will be much stronger if the purlins are 
notched above the principal rafters, than if they are 
framed into the side of the principals; for by this 
means, when any weight is applied in the middle of 
the purlin, it cannot bend, being confined by the other 
rafters; and if it do, the sides of the other rafters must 
needs bend along with it; consequently it has the 
strength of all the other rafters sideways added to it. 

This volume will be followed by at least three more, 
on the subject of Carpentry and Joining, one volume 
of which will be devoted to Framing and Heavy Tim¬ 
bered Work. The other two or more volumes will be 
devoted to fine joining and cabinet work, so far as the 
latter art comes within the purview of the joiners’ 
work. 

Each volume will, of course, be illustrated with dia¬ 
grams and working details. The matter will be taken 
from executed works and from every available source. 


MODERN CARPENTRY. 


YOL. II. 

ADVANCED SERIES. 


QUESTIONS. 

1. Give definition of the term 4 ‘ Solid Geometry.’ 7 

2. To what purposes may a knowledge of Solid 
Geometry be applied? 

3. Explain what is meant by the horizontal projec¬ 
tion of an object. 

4. Show by sketch and describe the vertical projec¬ 
tion (or elevation) and horizontal projection (or plan) 
of a table in the middle of an oblong room. 

376 


QUESTIONS 


377 


5. Explain what is meant by the elevation or vertical 
projection of an object. 

6. What is the term given to a line that represents 
the angle formed by the wall and floor of a room? 

7. Give an explanation of what is meant by the term 
“ perpendicular.” 

8. Give an explanation of what is meant by the term 
‘ ‘ vertical.’ ’ 

9. Give a sketch and show how to determine the posi¬ 
tion and length of a given straight line, parallel to one 
of the planes of the projection. 

10. Give a sketch and show how to determine the 
length of a given straight line which is oblique to both 
planes of projection. 

11. Show by sketch and describe how to find the 
points wherein the prolongation of a right line would 
meet the planes of projection, when projections of same 
right line are given. 

12. Show by sketch and describe how to find from 
the given projection of two lines that intersect each other 
in space, the angles which they make with each other. 

13. Show by sketch and describe how to determine 
from the projection of two lines which intersect each 
other in the projections, whether the lines cut each other 
in space or not. 

14. Show by sketch and describe how to find the 
angle made by a plane with the horizontal plane of pro¬ 
jection. 

15. When the traces of a plane and the projections of 
a point are given, show how to draw through the point 
a plane parallel to the given plane. 

16. When the traces of two planes which cut each 
other are given, show by sketches and describe how to 
find their intersections. 


378 


MODERN CARPENTRY 


17. When the traces of two intersecting planes are 
given, show by sketches and describe how to find the 
angle which the planes make between them. 

18. Show by sketches and describe how through a 
given point to draw a perpendicular to a given plane. 

19. Show by sketches and describe how through a 
given point to draw a plane perpendicular to a given 
right line. 

20. When a right line is given in projection and also 
the traces of a given plane, show by sketches and describe 
how to find the angle which the line makes with the 
plane. 

21. Show by sketch and. describe how to find the 
vertical projection of a regular tetrahedron when the 
horizontal projection is given. 

22. When a point is given in one of the projections of 
a tetrahedron, show by sketch and describe how to find 
the point on the other projection. 

23. When a tetrahedron, and the trace of a plane 
(perpendicular to one of the planes of projection) cut¬ 
ting it, by which it is truncated, are given, show by 
sketch and describe how to find the projection of the 
section. 

24. When the projections of a tetrahedron are given, 
show by sketch and describe how to find the projections 
when inclined to the horizontal plane in any degree. 

25. Show by sketch and describe how to construct 
the vertical and horizontal projections of a cube, the 
axes of which are perpendicular to the horizontal plane. 

26. Show by sketch and describe how to construct 
the projections of a regular octahedron, when one of its 
axes is perpendicular to either plane of projection. 

27. When one of the faces of an octahedron is given 
coincident with the horizontal plane of projection, show 


QUESTIONS 379 

by sketch and describe how to construct the projections 
of the solid. 

28. When the projection of one of the faces of a 
dodecahedron is given in the horizontal plane, show by 
sketches and describe how to construct its projections. 

29. When one of the faces of a dodecahedron is 
given, show by sketches and describe how to construct 
the projections of the solid, so that its axis may be 
perpendicular to the horizontal plane. 

30. Show by sketches and describe how to inscribe 
a tetrahedron, a hexahedron or cube, and a dodeca¬ 
hedron, in a given sphere. 

31. Show by sketches and describe how to find the 
vertical projection of the cylinder, when the horizontal 
projection, the axis of which is perpendicular to the 
horizontal plane, is given. 

32. When the traces of one oblique plane are given, 
show by sketch and describe how to determine the in¬ 
clination of the plane to both the horizontal position 
and the vertical position. 

33. Show by sketches and describe how to con¬ 
struct the projections of the cylinder, when the base 
and also the angles which the base makes with the 
planes of projection, are given. 

34. When a point in one of the projections of a cone 
is given, show by sketches and describe how to find it 
in the other projection. 

35. Show by sketch and describe a helix on a given 
cylinder. 

36. Show by sketch and describe a helix on a given 
cone. 

37. When the point in one of the projections of the 
sphere is given, show by sketches and describe how to 
find it in the other projection. 


380 


MODERN CARPENTRY 


38. When the projections of a regular tetrahedron 
are given, show by sketches and describe how to draw 
the section made by a plane perpendicular to the 
vertical plane and inclined to the horizontal plane. 

39. When the projections of a hexagonal pyramid 
are given, show by sketches and describe how to draw 
the section made by a plane perpendicular to the verti¬ 
cal plane and inclined to the horizontal plane. 

40. When the projections of an octagonal pyramid 
are given, show by sketches and describe how to draw 
the section made by a vertical plane. 

41. Show by sketches and describe how to draw the 
section of a cylinder by a plane oblique to the axis. 

42. In how many different ways may a cone be cut 
by ^a plane, and describe each separately, illustrated by 
sketch. 

43. Give sketch and describe how to draw the 
section of a cone made by a plane cutting both its 
sides, i. e., an ellipse. 

44. Give sketch and describe how to draw the section 
of a cone made by a plane parallel to one of its sides, 
a hyperboia. 

45. Give sketch and describe how to draw the sec¬ 
tion of a cone made by a plane parallel to the axis, i. e., 
a hyperbola. 

46. Give sketches and describe how to draw the sec¬ 
tion of a cuneoid made by a plane cutting both its sides. 

47. Give sketches and show how to describe the sec¬ 
tion of a cylinder made by a curve cutting the cylinder. 

48. Give sketches and show how to describe the sec¬ 
tion of a sphere. 

49. Show how to describe the section of an ellip¬ 
soid when a section through the fixed axis, and the 
position of the line of the required section, are given. 


QUESTIONS 


381 


50 . Show how to find the section of a cylindrical 
ring perpendicular to the plane passing through the 
axis of the ring, the line of section being given. 

51. Show how to describe the section of a solid of 
resolution, the generating curve of which is an ogee. 

52. Show how to find the section of a solid of resolu¬ 
tion, the generating curve of which is of a lancet form. 

53.. Show how to find the section of an ogee pyramid 
with a hexagonal base. 

54. When the projection of two equal cylinders 
which intersect at right angles are given, show how to 
find the projections of their intersections. 

55. Show how to construct the projections of twp 
unequal cylinders whose axes intersect each other ob¬ 
liquely. 

56. Show how to find the intersections of a sphere 
and a cylinder. 

57. Show how to construct the intersection of two 
right cones with circular bases. 

58. Explain the composition of the hexahedron or 

cube. 

59. Explain the composition of the octahedron. 

60. Explain the composition of the dodecahedron. 

61. Explain the composition of the icosahedron. 

62. Show wherein the operation consists in the de¬ 
velopment of a right pyramid of which the base and the 
height are given. 

63. Show how the development is found in an ob¬ 
lique pyramid. 

64. Show the results of the development of the faces 
and lines in a right prism. 

65. Show the results of the development of the faces 
and lines when a prism is inclined. 

66. Give the definition of a “cylinder.” 


382 


MODERN CARPENTRY 


67. Show how to find the covering of a right cylin¬ 
der. 

68. Show how to find the edge of the covering of a 
cylinder when it is oblique in regard to the sides. 

69. Show how to find the covering of a cylinder con¬ 
tained between two oblique parallel planes. 

70. Show how to find the covering of a semi-cylin- 
dric surface bounded by two curved lines. 

71. Show how to find the covering of the frustum of 
a cone, the section being made by a plane perpendicular 
to the axis. 

72. Show how to find the covering of a segmental 
dome. 

73. Show how to find the covering of a semi-circular 
dome. 

74. Show how to find the covering of a semi-circular 
dome when it is required to cover the dome horizontally- 

75. Show how to find the covering of an elliptic 
dome. 

76. Show how to find the covering boards of an 
ellipsoidal dome. 

77. Show how to find the covering of an ellipsoidal 
dome in gores. 

78. Show how to describe the covering of an ellip¬ 
soidal dome with horizontal boards of equal width. 

79. Show how to find the covering of an annular 
vault. 

80. Show how to find the covering of an ogee dome 
hexagonal in plan. 

81. Show how to find the exact stretch out or length 
of a straight line that shall equal the quadrant or a 
semi-circle. 

82. Show how to find the miters for intersecting 
straight and circular mouldings. 


QUESTIONS 


383 


83. Show how to find the form of a spring or solid 
moulding on any rake without the use of either ordi¬ 
nates or centers. 

84. Show by sketch and describe what is meant by 
the projection of a solid. 

85. Given the sides of a square bar which may be 
any length, the bar to stand perpendicular and pass 
through an inclined plank, show on the surface of 
plank the shape of a mortise that shall exactly fit the 
bar. 

86. Show by sketch and describe the mode of con¬ 
structing a bracketing ceiling under a roof. 

87. Give sketch of a six-paneled ceiling and describe 
the best way of making panelled ceilings. 

88. Show plan and section for a center suitable for 
a panelled ceiling and describe method of forming small 
ventilator for same. 

89. Show by sketch and describe how to find the 
angle bracket of a cornice^ for interior and exterior, 
otherwise re-entrant and salient, angles. 

90. Show by sketch and describe how to find the 
angle-bracket at the meeting of a concave wall with a 
straight wall. 

91. Show by sketch and describe the method of find¬ 
ing the angle-bracket at the meeting of coves of equal 
height but unequal projection. 

92. Show by sketch and describe the curb and ribs 
of a circular opening, cutting in on a sloped ceiling. 

93. Show by sketch and detail how to describe a 
spherical niche on a semi-circular plan. 

94. Show by sketch and detail how to describe a 
niche of which both the plan and elevation are seg¬ 
ments of a circle. 


384 


MODERN CARPENTRY 


95. Show by sketch and describe how to find the 
lengths and bevel of the ribs. 

96. Show by sketch and describe a niche semi-elliptic 
in plan and elevation. 

97. Show by sketch and describe how to draw the 
ribs of an irregular octagonal niche. 

99. Show by sketch in double curvature work and 
describe how to obtain to soffit mould for marking the 
veneer. 

100. Show by sketch and describe how to form the 
head. 

101. Show by sketch and describe how to obtain a 
.developed face mould. 

102. Show by sketch and describe how to find the 
mould for the cot bar. 

103. Show by sketch and describe how to find the 
mould for the radial bars. 

104. Show by sketch and describe how to make a 
frame-splayed lining with circular soffit. 

105. Show by sketch and describe how to obtain the 
face moulds for the soffit stiles. 

106. Show by sketch and describe how to apply the 
moulds. 

107. Show by sketch and describe a window with a 
splayed soffit and splayed jambs. 

108. Show by sketches and give description of how 
to enlarge and diminish mouldings. 

109. Show by sketch and describe method of finding 
the true section of an inclined moulding that is required 
to miter with a similar horizontal moulding at its lower 
end, as in pediments of doors and windows. 

110. What is meant by “sprung mouldings’*? 

111. Show by sketch and describe the mitering of 
straight and curved mouldings together. 


QUESTIONS 


385 


112. Show by sketch and describe how to obtain the 
section of a sash bar raking in plan. 

113. Give the different names applied to interior 
shutters. 

114. Give the different names applied to exterior 
shutters. 

115. Give a description of the folding or boxing shut¬ 
ters. 

116. Give sketch of a sectional elevation of a window 
fitted with boxing shutters. 

117. Give a description of sliding shutters. 

118. Give a description of balanced or lifting shut¬ 
ters. 

119. Give a description of rolling or spring shutters. 

120. Give description of the details in construction 
of windows. 

121. Give description of the joints of window 
sashes. 

122. Give description of “casement windows” in gen¬ 
eral. 

123. Give description of casement sashes opening 
inward. 

124. Show by sketch the elevation and vertical and 
horizontal sections of a window in a 14-inch brick wall 
fitted with a casement window having folding sashes 
to open inwards. 

125. Give description of casement sashes opening 
outwards, and sketches to illustrate same. 

126. Give description of details in the construction 
of a sash and frame window,'also sketches to illustrate 
same. 

127. Describe what is meant by a “double-hung 
sash and frame window.” 


386 


MODERN CARPENTRY 


128. Describe what is meant by a “single-hung sash 
and frame window.” 

129. Give description of a “Venetian window.” 

130. Give description of the hanging of vertical 
sliding sashes. 

131. Give description of a “bay window,” also show 
sketches of elevation and vertical section of same. 

132. Give description of a window with curved head, 
also sketch showing the outside and inside elevations. 

133. Give descriptions of the three ways how the 
arch may be formed when the head of the frame has 
to be curved. 

134. Give description of the construction of “store- 
windows. 9 ’ 

135. Give sketches showing the front elevation also 
vertical and horizontal sections of a store-window. 

136. Give sketch of part of a sash and frame win¬ 
dow, showing panelled and moulded jamb lining, etc. 

137. Give description of the “window linings,” of 
the “Architrave or Band Moulding” of the “Grounds” 
and “Window Board.” 

138. Give description of “window shutters.” 

139. Give details of the construction of “box shut¬ 
ters.” 

140. Give details of the construction of “sliding 
shutters . 9 ’ 

141. Give details as to the fitting of hinged sky¬ 
lights. 

142. Give details in the construction of “Dormer 
windows . 9 9 

143. Give details in the construction of “large 
skylights.” 

144. Give details in the construction of “lanter? 
lights.” 


QUESTIONS 387 

145. Give details in the construction of 4 ‘Lay- 
lights.’ J 

146. Give description of how “Greenhouses and 
Conservatories” are constructed. 

147. Give description of the construction of a “Bay 
window” with solid frame, and casement lights, also 
sketches showing plan, sections and elevations of same. 

148. Give description of the construction of a “cased 
frame bay window,” also sketches showing half plan, 
half inside elevation, and central vertical section of 
same. 

149. Give description of a “pivoted light” and its 
construction, also sketches of elevation and section of a 
pivot-liung sash. 

150. Give details of how to hang the sash of a 
“pivot-light.” 

151. Describe how to find the position to cut the 
beads for a “pivot-light.” 

152. Describe the construction of a “bull’s eye” 
frame with a pivoted sash, and give sketches of plan, 
and section of same. 

153. Describe the construction and laying out of 
a “circular louvre” and give sketches illustrating same. 

154. Give the names applied to the different kinds of 
doors. 

155. What are the essential qualities in the design 
and construction of doors? 

156. Describe “battened and battened frames and 
braced doors” and give sketches illustrating same. 

157. Describe the construction of the plain “bat¬ 
tened door” and give sketches of the “back elevation” 
and “vertical section” of same. 

158. Describe the construction of the “framed bat- 


388 


MODERN CARPENTRY 


tened and braced door” and give sketches of the “back 
elevation” and “vertical section” of same. 

159. How are “framed or panelled doors” distin¬ 
guished ? 

160. What is meant by the term “square and sunk”? 

161. What is meant by the term “moulded and 
square ’ ’ ? 

162. What is meant by the term “bead-flush”? 

163. What is meant by the term “bead-butt”? 

164. What is meant by the term “raised panel”? 

165. What is meant by the term “the chamfered 
panel ’ ’ ? 

166. What is meant by the term “raised and flat or 
raised and fielded?” 

167. What is meant by the term “raised, sunk, and 
fielded ’ ’ ? 

168. What is meant by the term “raised, sunk and 
moulded ’ ’ ? 

169. What is meant by the term, “stop-chamfered”? 

170. What is meant by the term “bolection mould¬ 
ed”? 

171. What is meant by the term “double bolection 
moulded ’ ’ ? 

172. Give the constructive memoranda for doors. 

173. Give sketch of a panelled door showing top and 
bottom rails, middle or lock rail, top and bottom munt- 
ing, stile, and top panels (one bead and flush) and (one 
square and sunk) and bottom panels (one moulded and 
square) and (one bead butt) also vertical section of 
door. 

174. Give sketches and describe how “bead flush” 
panels are commonly made. 

175. Describe construction of “bead butt” panels. 

176. Describe construction of “bolection mouldings.’* 


QUESTIONS 


389 


177. What are the usual dimensions given to ordi¬ 
nary good class dwelling house doors: Entrance doors, 
reception room doors, and bedroom doors, respectively? 

178. Give details in construction of interior doors. 

179. Give the names of usual kinds of timber em¬ 
ployed in the construction of doors, and to which class 
respectively. 

180. Give details as to the arrangement of members 
in the construction of doors. 

181. What are “double-margined doors” and when 
are they generally adopted? 

182. Describe the construction of “double-margined 
doors. ’ ’ 

183. Give sketches and describe how to “set out” 
the bevelled shoulders of the stile and rail of a “dimin¬ 
ished stile door. ’ ’ 

184. Give description of how to “set out” the rail. 

185. Give sketch of a “Gothic door” and describe 
its construction. 

186. Give sketches of “revolving doors” and de¬ 
scribe their construction. 

187. Give description of the construction of “pan¬ 
elled framing” and where adopted. 

188. Give description of ‘^superior doors” where two 
kinds of wood are used in their construction. 






INDEX 


A 

An Annular Vault . 128 

An Ogee Dome . 129 

Angle Brackets for Core. 153 

Angle Bracket for Curved and Straight Walls... 155 

Angle Bracket on a Rake. 157 

A Frame Splayed Lining. 175 

A Window with Splayed Soffit. 178 

Angle Sash Bars. 187 

Axle Pulleys . 213 

A Lantern Light. 233 

A Bay Window. 237 

A Cased Frame Bay Window. 240 

A Riveted Light. 243 

A Bull ’s-Eye Frame. 247 

A Segment Headed Frame. 286 

A Semi-Headed Frame. 287 

A Solid Frame .• •. 288 

Architrave Grounds. 294 

Architraves .. 295 

A Variety of Panel Doors.299-303 

B 

Bracketing for Cornices . 153 

Brackets for Re-entrant Angles. 154 

391 

























392 


INDEX 


Bracket for Curved Cornice . 155 

Boxing Shutters. 191 

Balancing Shutters . 193 

Bay Windows . 216 

Box Shutters. 225 

Barge Boards . 230 

Battered Doors . 252 

Brased Doors . 252 

Bead Flush Doors. 255 

Bead and Butt Doors. 256 

Bolection Moulded Doors. 257 

Bead Flush Panels. 259 

Bead Butt Panels. 259 

Bolection Mouldings . 261 

Bed-room Doors . 263 

Blind Screwing. 281 

Blind Nailing. 281 

Bending Ornaments. 345 

C 

Cube . 43 

Cube Projection . 44 

Cube and Octagon . 45 

Cone, Sphere, and Cylinder. 57 

Cylinder, Cone and Sphere. 57 

Conic Sections. 72 

Cones from Parabolas ... /. 75 

Cones and Hyperbolas. 76 

Cuneoids . 78 

Cylinder Sections . 78 

Cylindrical Rings . 82 

Cylinder and Sphere. 90 
































INDEX 


393 


Covering of Solids. 96 

Covering Prisms . 98 

Covering of Pyramids . 99 

Covering of Right Cylinders. 103 

Covering of Oblique Cylinders. 105 

Covering Semi-Cylindric Surfaces . 106 

Cone Coverings . 108 

Covering a Segmental Dome. 117 

Covering an Elliptic Dome. 122 

Covering an Ellipsoidal Dome. 123 

Covering an Annular Vault. 128 

Covering an Ogee Dome. 129 

Cutting Oblique Spouts or Timbers. 143 

Cutting Different Pitches. 144 

Ceilings in Wood or Stucco. 146 

Cove Brackets . *. 153 

Cornice Bracketing. 153 

Corner Bracket on a Rake. 157 

Curb Ribs for Circular Ceiling. 158 

Circular Splayed Work. 176 

Construction of Windows. 201 

Casement Windows . 204 

Cased Pulley Styles. 209 

Circular Head Windows. 220 

Corner Posts and Cross-Rails. 229 

Conservatories . 236 

Chamfered Doors. 256 

Constructive Memoranda . 257 

Common Doors . 264 

P> 

Diagram Showing Solid Geometry. 10 

Dodecahedron . 4 $ 

































394 


INDEX 


Dodecahedron Solids . 51 

Development of Coverings. . 97 

Diagram for Oblique Cuts.... . 145 

Double Semi-Circular Mitre. 160 

Double Curvature Work. 166 

Development of Conical Surfaces. 177 

Diminishing of Mouldings . 180 

Double Tenons . 203 

Double Sunk Rails . 208 

Double Hung Sashes. 214 

Double Bolection Moulded Doors. 257 

Double Margined Doors . 267 

Door and Window Grounds . 279 

Door and Jambs Complete. 298 

Deflection of Beams and Girders. 354 

E 

Ellipsoids . 86 

Ellipsoidal Coverings . 114 

Elliptical Dome Coverings. 120 

Elliptical Niche . ; .... 162 

Enlarging of Mouldings . 180 

Entrance Doors. 263 

External Doors. 264 

External Door Frames . 283 

F 

Finding Projection Points in a Cone. 60 

Finding Point of Projection in Sphere. 64 

Finding an Ellipsoid . 81 

Frustrum of Cones. Ill 





























INDEX 


395 


Finding the Stretchout of Curves. 131 

Framed Splayed Circular Soffit. 175 

Folding Shutters . 188 

Franking Sash Bars. 204 

Fixing Window Frames. 221 

Finials . 230 

Framed and Braced Doors. 253 

Framed Paneled Doors. 254 

Framed Grounds. 295 

Forces which Balance Each Other... 310 

G 

Given Points in Planes. 32 

Generating Curves. 84 

Gable Mitres . 140 

Grounds. 226 

Greenhouses and Conservatories. 236 

Grounds for Base . 291 

H 

Horizontal Traces.. 24 

Horizontal Projection . 28 

Helix in Cylinder . 62 

Helix in a Cone. 63 

Helical Curve on Sphere . 65 

Hexagonal Pyramid . 67 

How to Cut Oblique Spouts, etc. 143 

Hook Point . 207 

Hinged Skylights . 227 

Heavy Architraves..*. 296 




























396 


INDEX 


I 

Intersecting Lines . 18 

Intersection of Curved Surfaces . 86 

Intersection of Cylinders. 87 

Intersecting a Sphere and Cylinder. 90 

Intersection of Cones ... .. 91 

Irregular Mitres. 133 

Interior Shutters . 188 

Inclined Forces. 323 

J 

Joints of Sashes. 203 

L 

Lancet Curves. 84 

Large Panels for Ceilings . 151 

Long Shutters . 189 

Lifting Shutters ..• •. 193 

Linings . 223 

Large Sky-Lights. 231 

Lantern Lights . 231 

Laying Out a Circular Louvre. 248 

Louvre Boards . 249 

Louvre Gains. 250 

Loaded Beams . 331 

M 

Mitres for Irregular Forms. 133 

Mitres on a Rake • •. 135 

Mitring Irregular Mouldings . 138 

Mitring Sprung Mouldings . 185 



























INDEX 


397 


Mitring Curved and Straight Mouldings. 185 

Mullions . 215 

Moulded and Square Doors. 255 

Moulded Grounds. 294 

Mechanics of Carpentry. 305 

Methods of Arranging Beams.• • 344 

N 

Niches .159 

Niche on Elliptical Plane. 164 

0 

Octagon Diagram. 13 

Octagon and Cube . 44 

Octahedron . 46 

Octahedron Projection . 47 

Oblique Planes. 57 

Octagonal Pyramid. 69 

Oblique Axes . 70 

Ogee Pyramids . 85 

Oblique Cuts ...• •.143 

On Mitres. 159 

Octagonal Mitre .»1.• •.163 

Outside and Inside Linings. 209 

Other Paneled Framing. 278 

P 

Preface . ^ 

Part I. ^ 

Points, Lines and Planes .. 16 

Planes and Projections... 22 

Perspective Lines and Traces. 20 






























398 


INDEX 


Projection of Cylinder. 

Pyramid . 

Prisms . 

Pyramids Developed . 

Practical Solutions. 

Projection of Solids . 

Paneled Ceilings, Wood or Stucco 

Parting Strips. 

Pocket Piece . 

Position to Cut Beads. 

Planted Moulded Doors. 

Projections of Doors. 

Plain Jamb Linings . 

Plinth Blocks. 

Panel Doors in Various Styles.... 

Polygon of Forces. 

Parallel Forces . 

Pulleys . 


R 

Right Lines . 

Regular Polyhedrons. 

Raking Moulding Mitres . 

Raking Bracket for Corner. 

Ribs for Circular Arch and Ceiling 

Ribs for Octagonal Niche. 

Regular Octagonal Niche. 

Ribs for Irregular Octagonal Niche 

Radial Sash Bars. 

Raking Mouldings . 

Raking Sash Bars. 

Rolling Shutters. 

































INDEX ' 


399 


Raised Panel Door. 256 

Raised and Flat Door. 256 

Raised, Sunk and Moulded Door. 256 

Raised, Sunk and Moulded. 257 

Reception Room Doors. 263 

Resistant of Two or More Forces. 306 

Reciprocal Diagrams. 332 

Revolving Doors. 278 

S 

Solid Geometry. 9 

Straight-Sided Solids . 36 

Sphere and Tetrahedron. 55 

Sphere, Cone and Cylinder . 57 

Sections of Solids. 56 

Sections of Cones. 76 

Sections of Cones from Parabolas. 75 

Sections of Cones from Hyperbolas.;. 76 

Sections of Cuneoids. 78 

Sections of Cylinder. 78 

Solids of Revolution . 83 

Sphere and Cylinder. 90 

Semi-Cylindric Coverings. 166 

Spheres and Ellipsoids . 113 

Spherical Niche. 1^9 

Semi-Circular Niche. 166 

Segmental Niche. 161 

Semi-Elliptical Niche. 163 

Segment Heads . 1^6 

Staving for Circular Heads, .. 170 

Sash Bars for Circular Windows. 173 

Splayed Work Circulars. 176 
































400 


INDEX 


Sprung Mouldings . 185 

Shutters . 188 

Sliding Shutters . 193 

Spring Shutters .. 197 

Size and Position of Windows. 201 

Sashes . 201 

Sashes Opening Inward ... 205 

Sashe^ Opening Outward. 208 

Sash and Frame Window. 209 

Sashes and Window Frames. 209 

Sashes of Various Kinds. 212 

Single Hung Sashes. 214 

Sliding Sashes.215 

Shop Windows . 221 

Sliding Shutters . 226 

Sash Frame . 228 

Side-Lights . 233 

Square and Sunk Doors. 254 

Stop Chamfered Doors. 257 

Solid Moulded Doors. 258 

Superior Internal Doors . 264 

Square Shoulder Rail. 271 

Superior Doors . 279 

Secret Fastening . 281 

Solid Door Frames . 283 

Semi-Circular Frame . 288 

Skeleton Frame ... 289 

Set of Linings. 289 

Stress Diagrams for Trusses. 337 

Strength of Wooden Beams. 342 

Stiffners of Beams. 342 

Specific Gravity . 358 

Strength of Timber. 359 



































INDEX 


401 


T 

To Determine Length of Oblique Lines. 16 

To Determine Various Angles. 21 

Traces of Intersecting Planes. 25 

Tetrahedron ...... 37 

Tetrahedron Projection . 40 

To Find Inclination of Two Faces. 49 

To Describe a Helix. 64 

The Cone . 71 

To Describe the Section of a Sphere. 80 

To Describe an Ellipsoid. 81 

To Intersect Two Unequal Cylinders. 88 

To Find Covering of Frustrum of Cone. 112 

To Find the Covering of an Elliptic Dome. 122 

To Find Angle Bracket for Cornice. 154 

To Obtain Soffit Mould. 166 

To Form a Head . 168 

To Obtain a Developed Face Mould. 172 

To Find Mould for a “Cot Bar”.. 173 

To Find Mould for Radial Bars. 173 

To Obtain Face Moulds .. 176 

To Apply the Moulds. 177 

The Development of Conical Surfaces. 177 

To Diminish a Moulding. 181 

Throating . 205 

To Hang the Sash. 245 

To Find Position to Cut Beads. 245 

The Construction of Doors. 251 

To Set Out Gunstock Stiles. 270 

To Set Out Rail for Gunstock Stile. 271 

The Nature of Force... 306 

The Parallelogram of Forces. 308 

































402 


INDEX 


Triangle of Forces. 312 

Transverse Strength of Wooden Beams. 349 


V 


Vertical and Horizontal Projections. 17 

Vertical Projection . 31 

Vertical Projection of Dodecahedron. 53 

Veneering Soffits. 170 

Venetian Windows .215 

Vertical Sashes . 215 


W 


Windows Generally . 201 

Window Frames. 204 

Window and Sash-Frames . 210 

Windows with Curved Heads. 219 

Windows for Shops. 221 

Wedges . 221 

Window Linings. 223 

Window Shutters .225 

Window Grounds ...\. 279 

Questions for Students.376 



N 58 






























































